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In the figure shown the switch S1 remains connected for long time and the switch S2 remains opened. Now the switch S2 is closed. Assuming that ε = 10 volt and L = 1H, the magnitude of rate of change of current is 2X (in Amp/sec.) in the inductor just after the switch S2 is closed then find X .
Correct answer is '5'. Can you explain this answer?
In the figure shown the switch S1 remains connected for long time and the switch S2 remains opened. Now the switch S2 is closed. Assuming that ε = 10 volt and L = 1H, the magnitude of rate of change of current is 2X (in Amp/sec.) in the inductor just after the switch S2 is closed then find X .
|
Vp Classes answered |
Amphoteric oxides, such as aluminium oxide, are soluble both in strongly acidic and in strongly basic
solutions :
Dissolution of Al(OH)3 in excess base is just a special case of the effect of complex-ion formation on solubility. Al(OH)3 dissolves because excess OH¯ ions convert it to the soluble complex ion Al(OH)4¯ (aluminate ion).The effect of pH on the solubility of Al(OH)3 is shown in figure. 
Other examples of amphoteric hydroxides include Zn(OH)2, Cr(OH)3, Sn(OH)2 and Pb(OH)2, which react with
excess OH- ions to form the soluble complex ions Zn(OH)42- (zincate ion),Cr(OH)4- (chromite ion),Sn(OH)3¯(stannite ion), and Pb(OH)3¯ (plumbite ion), respectively. By contrast, basic hydroxides, such as Mn(OH)2, Fe(OH)2,and Fe(OH)3, dissolve in strong acid but not in strong base.Zn(OH)2 is a amphoteric hydroxide and is involved in the following two equilibria in aqueous solutions
At what pH the solubility of Zn(OH)2 be minimum ?- a)4
- b)10
- c)6
- d)8
Correct answer is option 'B'. Can you explain this answer?
Amphoteric oxides, such as aluminium oxide, are soluble both in strongly acidic and in strongly basic
solutions :
solutions :
Dissolution of Al(OH)3 in excess base is just a special case of the effect of complex-ion formation on solubility. Al(OH)3 dissolves because excess OH¯ ions convert it to the soluble complex ion Al(OH)4¯ (aluminate ion).The effect of pH on the solubility of Al(OH)3 is shown in figure.
Other examples of amphoteric hydroxides include Zn(OH)2, Cr(OH)3, Sn(OH)2 and Pb(OH)2, which react with
excess OH- ions to form the soluble complex ions Zn(OH)42- (zincate ion),Cr(OH)4- (chromite ion),Sn(OH)3¯(stannite ion), and Pb(OH)3¯ (plumbite ion), respectively. By contrast, basic hydroxides, such as Mn(OH)2, Fe(OH)2,and Fe(OH)3, dissolve in strong acid but not in strong base.
excess OH- ions to form the soluble complex ions Zn(OH)42- (zincate ion),Cr(OH)4- (chromite ion),Sn(OH)3¯(stannite ion), and Pb(OH)3¯ (plumbite ion), respectively. By contrast, basic hydroxides, such as Mn(OH)2, Fe(OH)2,and Fe(OH)3, dissolve in strong acid but not in strong base.
Zn(OH)2 is a amphoteric hydroxide and is involved in the following two equilibria in aqueous solutions
At what pH the solubility of Zn(OH)2 be minimum ?
a)
4
b)
10
c)
6
d)
8
|
Krishna Iyer answered |
Let solubility of Zn(OH)2 = s, some of will go in Zn2+ form and some in complex |Zn(OH)4|2-, of 
= 10-x M,
........(1)
[S1 + S2 total solubility]
So, S1 S2 = 1.2 x 1.2 10-18
Now, we want S = S1 + S2 to be minimum we will have S1 = S2
So, = S1 = S2 = 1.2 x 10-9 M
Hence from 1" equation we get
So,
Hence x = 10
= 10-x M, ........(1) [S1 + S2 total solubility]So, S1 S2 = 1.2 x 1.2 10-18
Now, we want S = S1 + S2 to be minimum we will have S1 = S2
So, = S1 = S2 = 1.2 x 10-9 M
Hence from 1" equation we get
So,
Hence x = 10
A rod of length l having uniformly distributed charge Q is rotated about one end with constant frequency ' f '. Its magnetic moment. - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A rod of length l having uniformly distributed charge Q is rotated about one end with constant frequency ' f '. Its magnetic moment.
a)
b)
c)
d)
|
Vijay Kumar answered |
Charge on the differential element dx,
equivalent current di = f dq
∴ magnetic moment of this element
equivalent current di = f dq
∴ magnetic moment of this element
dμ = (di) NA (N = 1)
⇒
⇒
In the figure shown a parallel plate capacitor has a dielectric of width d/2 and dielectric constant K = 2. The other dimensions of the dielectric are same as that of the plates. The plates P1 and P2 of the capacitor have area 'A' each. The energy of the capacitor is : 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
In the figure shown a parallel plate capacitor has a dielectric of width d/2 and dielectric constant K = 2. The other dimensions of the dielectric are same as that of the plates. The plates P1 and P2 of the capacitor have area 'A' each. The energy of the capacitor is :
a)
b)
c)
d)
|
Lalit Yadav answered |
A water tank stands on the roof of a building as shown. Then the value of 'h' for which the distance covered by the water 'x' is greatest is - 
- a)0.5 m
- b)0.67 m
- c)1 m
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
A water tank stands on the roof of a building as shown. Then the value of 'h' for which the distance covered by the water 'x' is greatest is -
a)
0.5 m
b)
0.67 m
c)
1 m
d)
none of these
|
Rohit Jain answered |
the roots of x are (0,4) and the maximum of x is at h = 2.
The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval.
Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint ,lower the range.Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m.
The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is / are incorrect (in the low pressure region):
- a)H2 and He show positive deviation from ideal gas equation.
- b)CO, CH4 and O2 show negative deviation from ideal gas equation.
- c)H2 and He show negative deviation while CO2, CH4 and O2 show positive deviation.
- d)H2 and He are less compressible than that of an ideal gas while CO2, CH4 and O2 more compressible than that of ideal gas.
Correct answer is option 'C'. Can you explain this answer?
The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is / are incorrect (in the low pressure region):
a)
H2 and He show positive deviation from ideal gas equation.
b)
CO, CH4 and O2 show negative deviation from ideal gas equation.
c)
H2 and He show negative deviation while CO2, CH4 and O2 show positive deviation.
d)
H2 and He are less compressible than that of an ideal gas while CO2, CH4 and O2 more compressible than that of ideal gas.
|
Yash Patel answered |
If Z > 1 positive deviation
Z < 1 negative deviation
Z < 1 negative deviation
A bell rings at an interval of every 2 minutes, second bell rings at every 5 minutes, 3rd bell rings at every 6 minutes and 4th ring in every 8 minutes. In a span of 8 hours, how many times all 4 bells be ringing simultaneously- a)60
- b)32
- c)16
- d)4
Correct answer is option 'D'. Can you explain this answer?
A bell rings at an interval of every 2 minutes, second bell rings at every 5 minutes, 3rd bell rings at every 6 minutes and 4th ring in every 8 minutes. In a span of 8 hours, how many times all 4 bells be ringing simultaneously
a)
60
b)
32
c)
16
d)
4
|
Anaya Patel answered |
Product of 22 integers is equal to 1, then their sum cannot be- a)0
- b)1
- c)2
- d)4
Correct answer is option 'A,B,D'. Can you explain this answer?
Product of 22 integers is equal to 1, then their sum cannot be
a)
0
b)
1
c)
2
d)
4
|
Shraddha Sharma answered |
Only integers which can satisfy this condition are -1 & +1. taking 10 times -1 and 12times +1 we will get the product 1. and sum will result in 2. check other cases similarly.
A single circular loop of wire with radius 0.02 m carries a current of 8.0 A. It is placed at the centre of a solenoid that has length 0.65 m, radius 0.080 m and 1300 turns. 
- a)The value of the current in the solenoid so that the magnetic field at the centre of the loop becomes zero, is equal to 44 mA.
- b)The value of the current in the solenoid so that the magnetic field at the centre of the loop becomes zero, is equal to 100 mA.
- c)The magnitude of the total magnetic field at the centre of the loop (due to both the loop and the solenoid) if the current in the loop is reversed in direction from that needed to make the total field equal to zero tesla, is 8p x 10–5 T.
- d)The magnitude of the total magnetic field at the centre of the loop (due to both the loop and the solenoid) if the current in the loop is reversed in direction from that needed to make the total field equal to zero tesla, is 16p x 10–5 T.
Correct answer is option 'B,D'. Can you explain this answer?
A single circular loop of wire with radius 0.02 m carries a current of 8.0 A. It is placed at the centre of a solenoid that has length 0.65 m, radius 0.080 m and 1300 turns.
a)
The value of the current in the solenoid so that the magnetic field at the centre of the loop becomes zero, is equal to 44 mA.
b)
The value of the current in the solenoid so that the magnetic field at the centre of the loop becomes zero, is equal to 100 mA.
c)
The magnitude of the total magnetic field at the centre of the loop (due to both the loop and the solenoid) if the current in the loop is reversed in direction from that needed to make the total field equal to zero tesla, is 8p x 10–5 T.
d)
The magnitude of the total magnetic field at the centre of the loop (due to both the loop and the solenoid) if the current in the loop is reversed in direction from that needed to make the total field equal to zero tesla, is 16p x 10–5 T.
|
Neha Joshi answered |
A painter is applying force himself to raise him and the box with an cceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s2, then : 
- a)tension in the rope is 1125 N
- b)tension in the rope is 2250 N
- c)force of contact between the painter and the floor is 375 N
- d)none of these
Correct answer is option 'A,C'. Can you explain this answer?
A painter is applying force himself to raise him and the box with an cceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s2, then :
a)
tension in the rope is 1125 N
b)
tension in the rope is 2250 N
c)
force of contact between the painter and the floor is 375 N
d)
none of these
|
Cstoppers Instructors answered |
For the whole system, 2T – 1500 = 150 x 5 ⇒ T = 1125 N
For the person, T – 1000 + N = 100 x 5 N = 1500 – 1125 = 375 N
For the person, T – 1000 + N = 100 x 5 N = 1500 – 1125 = 375 N
In a sample of H-atom electrons make transition from 5th excited state to ground state, producing all possible types of photons, then number of lines in infrared region are - a)4
- b)5
- c)6
- d)3
Correct answer is option 'C'. Can you explain this answer?
In a sample of H-atom electrons make transition from 5th excited state to ground state, producing all possible types of photons, then number of lines in infrared region are
a)
4
b)
5
c)
6
d)
3
|
|
Krishna Iyer answered |
5 lines belong to ultra violet region and 4 lines are in visible region and rest are in infrared region.
The familiar brown ring test for nitrates depends on the ability of Fe2+ to reduce nitrates to nitric oxide, which reacts with Fe2+ to form a brown coloured complex [Fe(H2O)5NO+]SO4. What is the oxidation state of iron in the complex. (Only write the number 1, 2, 3, ......)
Correct answer is '1433'. Can you explain this answer?
The familiar brown ring test for nitrates depends on the ability of Fe2+ to reduce nitrates to nitric oxide, which reacts with Fe2+ to form a brown coloured complex [Fe(H2O)5NO+]SO4. What is the oxidation state of iron in the complex. (Only write the number 1, 2, 3, ......)
|
|
Amit Kumar answered |
The string of a step rolling wheel is pulled by applying force F with different lines of action in two situations as shown. The wheel starts rolling without slipping due to application of the force : 
- a)The wheel rolls to the right in situation I and to the left in situation II.
- b)The wheel rolls to the left in situation I and to the right in situation II
- c)The wheel rolls to the right in both situations.
- d)The wheel rolls to the left in both situations.
Correct answer is option 'A'. Can you explain this answer?
The string of a step rolling wheel is pulled by applying force F with different lines of action in two situations as shown. The wheel starts rolling without slipping due to application of the force :
a)
The wheel rolls to the right in situation I and to the left in situation II.
b)
The wheel rolls to the left in situation I and to the right in situation II
c)
The wheel rolls to the right in both situations.
d)
The wheel rolls to the left in both situations.
|
|
Krishna Iyer answered |
(Pure rotation about instantaneous point of contact)
Note : If line action passes through point of contact, it only spins.
Let ‘P’ be a point which does not lie outside the triangle ABC, A = (3, 2), B = (0, 0), C = (0, 4) which satisfy d (P, A) < maximum {d (P, B), d (P, C)} then maximum distance of P from side BC, where d (P, A) gives the distance between P & A, is - a)3/4
- b)4/3
- c)3
- d)0
Correct answer is option 'C'. Can you explain this answer?
Let ‘P’ be a point which does not lie outside the triangle ABC, A = (3, 2), B = (0, 0), C = (0, 4) which satisfy d (P, A) < maximum {d (P, B), d (P, C)} then maximum distance of P from side BC, where d (P, A) gives the distance between P & A, is
a)
3/4
b)
4/3
c)
3
d)
0
|
Sarita Yadav answered |
Two roads OA and OB intersect at an angle of 60º. A car driver approaches O from A where OA = 800m at a uniform speed of 20m/s. Simultaneously another car moves from O towards B at a uniform speed of 25m/s. If t is time when two cars are closest, find t.- a)1
- b)2
- c)1040/61
- d)3/2
Correct answer is option 'C'. Can you explain this answer?
Two roads OA and OB intersect at an angle of 60º. A car driver approaches O from A where OA = 800m at a uniform speed of 20m/s. Simultaneously another car moves from O towards B at a uniform speed of 25m/s. If t is time when two cars are closest, find t.
a)
1
b)
2
c)
1040/61
d)
3/2
|
|
Vijay Kumar answered |
Let the distance between two cars after t seconds be s. Distance covered by first car in t second is 20t meter, so its distance from O will be (800 – 20t) meter and distance covered by second car will be 25t meter.
Figure shows a uniformly charged hemispherical shell. The direction of electric field at point p, that is off-centre (but in the plane of the largest circle of the hemisphere), will be along 
- a)pa
- b)pb
- c)pc
- d)pd
Correct answer is option 'B'. Can you explain this answer?
Figure shows a uniformly charged hemispherical shell. The direction of electric field at point p, that is off-centre (but in the plane of the largest circle of the hemisphere), will be along
a)
pa
b)
pb
c)
pc
d)
pd
|
Pioneer Academy answered |
Let electric field at point. 'p' has both x and y component.
So similar electric field will be, for other hemisphere (upper half).
Now lets overlap both.
So similar electric field will be, for other hemisphere (upper half).
Now lets overlap both.
(Enet)p = 2 Ex and it should be zero (as E inside a full shell = 0).
So Ex = 0, So electric field at 'p' is purely in y direction.
The intensity received at the focus of the lens is I when no glass slab has been placed in front of the slit. Both the slits are of the same dimension and the plane wavefront incident perpendicularly on them, has wavelength λ. On placing the glass slab, the intensity reduces to 3I/4 at the focus. Find out the minimum thickness of the glass slab (in Å) if its refractive index is 3/2. Given λ = 6933Å, µ = 1.5.
Correct answer is '2311'. Can you explain this answer?
The intensity received at the focus of the lens is I when no glass slab has been placed in front of the slit. Both the slits are of the same dimension and the plane wavefront incident perpendicularly on them, has wavelength λ. On placing the glass slab, the intensity reduces to 3I/4 at the focus. Find out the minimum thickness of the glass slab (in Å) if its refractive index is 3/2. Given λ = 6933Å, µ = 1.5.
|
|
Krishna Iyer answered |
A is a thin walled sphere at rest made up of glass. The radius of the sphere is 1m and it is filled with a
transparent liquid of refractive index m. S is the luminous source moving directly towards a plane
vertical mirror M. A fish in the liquid is moving towards S. The eye of the fish and S are collinear &
perpendicular to mirror M. At the moment S is 3m away from the centre of the sphere, fish observes
that image of S due to reflection is moving with speed of 13m/s. If speed of the fish relative to the
sphere is 10m/s and m = 1.5 then find the speed of the source at that instant. The system is placed
in air.
Correct answer is '8'. Can you explain this answer?
A is a thin walled sphere at rest made up of glass. The radius of the sphere is 1m and it is filled with a
transparent liquid of refractive index m. S is the luminous source moving directly towards a plane
vertical mirror M. A fish in the liquid is moving towards S. The eye of the fish and S are collinear &
perpendicular to mirror M. At the moment S is 3m away from the centre of the sphere, fish observes
that image of S due to reflection is moving with speed of 13m/s. If speed of the fish relative to the
sphere is 10m/s and m = 1.5 then find the speed of the source at that instant. The system is placed
in air.
transparent liquid of refractive index m. S is the luminous source moving directly towards a plane
vertical mirror M. A fish in the liquid is moving towards S. The eye of the fish and S are collinear &
perpendicular to mirror M. At the moment S is 3m away from the centre of the sphere, fish observes
that image of S due to reflection is moving with speed of 13m/s. If speed of the fish relative to the
sphere is 10m/s and m = 1.5 then find the speed of the source at that instant. The system is placed
in air.
|
Impact Learning answered |
Which of the following statements is/are correct ? - a)Out of trimethylamine and trimethylphosphine, trimethylamine has higher dipole moment ?
- b)Out of (SiH3)2O and (CH3)2O, (SiH3)2O is more basic.
- c)C–C bond length (in pm) in C2 molecule is greater than O–O bond length in O2 molecule.
- d)N(SiMe3)3 and BF3 molecules are isostructural.
Correct answer is option 'A,C,D'. Can you explain this answer?
Which of the following statements is/are correct ?
a)
Out of trimethylamine and trimethylphosphine, trimethylamine has higher dipole moment ?
b)
Out of (SiH3)2O and (CH3)2O, (SiH3)2O is more basic.
c)
C–C bond length (in pm) in C2 molecule is greater than O–O bond length in O2 molecule.
d)
N(SiMe3)3 and BF3 molecules are isostructural.
|
Lavanya Menon answered |
(A) Nitrogen is more electronegative than phosphorus.
So, dipole moment of trimethylamine is greater than trimethy phosphine.
In trisilyl ether the lone pair of electron on oxygen atom is less easily available for donation because of pπ-dπ
delocalisation due to presence of the vacant d-orbital with Si. This however is not possible with carbon in CH3–O–CH3 due to the absence of d-orbital making it more basic.
(C) Bond order of C2
and O2 are same i.e., 2. In C2 molecules both bonds are π-bonds whereas, there is one σ and one π-bond in O2 molecule
C2 = 131 pm ; O2 = 121 pm.
delocalisation due to presence of the vacant d-orbital with Si. This however is not possible with carbon in CH3–O–CH3 due to the absence of d-orbital making it more basic.
(C) Bond order of C2
and O2 are same i.e., 2. In C2 molecules both bonds are π-bonds whereas, there is one σ and one π-bond in O2 molecule
C2 = 131 pm ; O2 = 121 pm.
An object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between object and its image will be :- a)increasing
- b)decreasing
- c)first increases then decreases
- d)first decreases and then increases
Correct answer is option 'B'. Can you explain this answer?
An object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between object and its image will be :
a)
increasing
b)
decreasing
c)
first increases then decreases
d)
first decreases and then increases
|
Praveen Kumar answered |
Suppose m boys and m girls take their seats randomly round a circle. The probability of their sitting is (2mCm )–1 , when- a)no two boys sit together
- b)no two girls sit together
- c)boys and girls sit alternately
- d)all the boys sit together
Correct answer is option 'A,B,C'. Can you explain this answer?
Suppose m boys and m girls take their seats randomly round a circle. The probability of their sitting is (2mCm )–1 , when
a)
no two boys sit together
b)
no two girls sit together
c)
boys and girls sit alternately
d)
all the boys sit together
|
|
Yash Patel answered |
The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation 
When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl– and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively.Q. What is the molar conductance of NaCl at infinite dilution ?- a)147 ohm–1 cm2 mole–1
- b)107 ohm–1 cm2 mole–1
- c)127 ohm–1 cm2 mole–1
- d)157 ohm–1 cm2 mole–1
Correct answer is option 'C'. Can you explain this answer?
The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation
When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl– and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively.
Q. What is the molar conductance of NaCl at infinite dilution ?
a)
147 ohm–1 cm2 mole–1
b)
107 ohm–1 cm2 mole–1
c)
127 ohm–1 cm2 mole–1
d)
157 ohm–1 cm2 mole–1
|
|
Anaya Patel answered |
Statement-1 : Two waves moving in a uniform string having uniform tension cannot have different velocities.Statement-2 : Elastic and inertial properties of string are same for all waves in same string. Moreover speed of wave in a string depends on its elastic and inertial properties only.- a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
- b)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
- c)Statement-1 is True, Statement-2 is False
- d)Statement-1 is False, Statement-2 is True
Correct answer is option 'D'. Can you explain this answer?
Statement-1 : Two waves moving in a uniform string having uniform tension cannot have different velocities.
Statement-2 : Elastic and inertial properties of string are same for all waves in same string. Moreover speed of wave in a string depends on its elastic and inertial properties only.
a)
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
b)
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c)
Statement-1 is True, Statement-2 is False
d)
Statement-1 is False, Statement-2 is True
|
|
Yash Patel answered |
Two waves moving in uniform string with uniform tension shall have same speed and may be moving in opposite directions. Hence both waves may have velocities in opposite direction. Hence statement-1 is false.
A current of 2.68 A is passed for one hour through an aqueous solution of CuSO4 using copper electrodes . Select the correct statement(s) from the following :- a)increase in mass of cathode = 3.174 g
- b)decrease in mass of anode = 3.174 g
- c)no change in masses of electrodes
- d)the ratio between the change of masses of cathode and anode is 1 : 2 .
Correct answer is option 'A,B'. Can you explain this answer?
A current of 2.68 A is passed for one hour through an aqueous solution of CuSO4 using copper electrodes . Select the correct statement(s) from the following :
a)
increase in mass of cathode = 3.174 g
b)
decrease in mass of anode = 3.174 g
c)
no change in masses of electrodes
d)
the ratio between the change of masses of cathode and anode is 1 : 2 .
|
|
Anaya Patel answered |
Increase in mass of cathode = decrease in mass of Anode =
Statement-1 : A point object moves near the principal axis of a fixed spherical mirror along a straight line. Then the image formed by the spherical mirror also moves along a straight line.Statement-2 : For an incident ray on a fixed spherical mirror there is a fixed reflected ray. If a point object moves along this incident ray, its image will always lie on the given reflected ray. Further an incident ray may be drawn from the moving point object in its direction of velocity towards the mirror.- a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
- b)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
- c)Statement-1 is True, Statement-2 is False
- d)Statement-1 is False, Statement-2 is True
Correct answer is option 'A'. Can you explain this answer?
Statement-1 : A point object moves near the principal axis of a fixed spherical mirror along a straight line. Then the image formed by the spherical mirror also moves along a straight line.
Statement-2 : For an incident ray on a fixed spherical mirror there is a fixed reflected ray. If a point object moves along this incident ray, its image will always lie on the given reflected ray. Further an incident ray may be drawn from the moving point object in its direction of velocity towards the mirror.
a)
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
b)
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c)
Statement-1 is True, Statement-2 is False
d)
Statement-1 is False, Statement-2 is True
|
Geetika Shah answered |
Draw an incident ray on the mirror and trace the corresponding reflected ray. If a point object moves along this ray,its image will always lies on the traced reflected ray. Hence when a point object moves near the principal axisof a fixed spherical mirror along a straight line, then its image formed by the spherical mirror also moves along
a straight line.Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
a straight line.Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
Match the statements in column-I with the statements in column-II.
Column-I Column-II
(A) A tight string is fixed at both ends and (p) At the middle, antinode is formed
sustaining standing wave in odd harmonic
(B) A tight string is fixed at one end and (q) At the middle, node is formed
free at the other end in even harmonic
(C) Standing wave is formed in an open organ (r) At the middle, neither node nor
pipe. End correction is not negligible. antinode is formed
(D) Standing wave is formed in a closed (s) Phase difference between SHMs of any
organ pipe. End correction is not negligible. two particles will be either p or zero. (t) The displacement of the particle in the middle is always non zero.- a)(A) p,q,s (B) r,s (C) s (D) r,s
- b)(A) r,q (B) s (C) p,q,r (D) p
- c)(A) r (B) p,q (C) r,p (D) p,q
- d)(A) r (B) q (C) p,q (D) r
Correct answer is option 'A'. Can you explain this answer?
Match the statements in column-I with the statements in column-II.
Column-I Column-II
(A) A tight string is fixed at both ends and (p) At the middle, antinode is formed
sustaining standing wave in odd harmonic
(B) A tight string is fixed at one end and (q) At the middle, node is formed
free at the other end in even harmonic
(C) Standing wave is formed in an open organ (r) At the middle, neither node nor
pipe. End correction is not negligible. antinode is formed
(D) Standing wave is formed in a closed (s) Phase difference between SHMs of any
organ pipe. End correction is not negligible. two particles will be either p or zero.
Column-I Column-II
(A) A tight string is fixed at both ends and (p) At the middle, antinode is formed
sustaining standing wave in odd harmonic
(B) A tight string is fixed at one end and (q) At the middle, node is formed
free at the other end in even harmonic
(C) Standing wave is formed in an open organ (r) At the middle, neither node nor
pipe. End correction is not negligible. antinode is formed
(D) Standing wave is formed in a closed (s) Phase difference between SHMs of any
organ pipe. End correction is not negligible. two particles will be either p or zero.
(t) The displacement of the particle in the middle is always non zero.
a)
(A) p,q,s (B) r,s (C) s (D) r,s
b)
(A) r,q (B) s (C) p,q,r (D) p
c)
(A) r (B) p,q (C) r,p (D) p,q
d)
(A) r (B) q (C) p,q (D) r
|
Aadhar Academy answered |
(A) Number of loops (of length λ/2) will be even or odd and node or antinode will respectively be formed at the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent
loops will be π.
(B) and (D) Number of loops will not be integral. Hence neither a node nor an antinode will be formed in in the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent
loops will be π.
(C) Number of loops (of length λ/2) will be even or odd and antinode or node will respectively be formed at the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be π ..
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent
loops will be π.
(B) and (D) Number of loops will not be integral. Hence neither a node nor an antinode will be formed in in the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent
loops will be π.
(C) Number of loops (of length λ/2) will be even or odd and antinode or node will respectively be formed at the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be π ..
In the figure shown a small block ‘B’ of mass ‘m’ is released from the top of a smooth movable wedge ‘A’ of the same mass ‘m’. ‘B’ ascends another movable smooth wedge ‘C’ of the same mass. Neglecting friction any where the maximum height attained by ‘B’ on ‘C’ is h/2x Find the value of x .
Correct answer is '2'. Can you explain this answer?
In the figure shown a small block ‘B’ of mass ‘m’ is released from the top of a smooth movable wedge ‘A’ of the same mass ‘m’. ‘B’ ascends another movable smooth wedge ‘C’ of the same mass. Neglecting friction any where the maximum height attained by ‘B’ on ‘C’ is h/2x Find the value of x .
|
|
Dev Kumar answered |
If a variable tangent of circle x2 + y2 = 1 intersects the ellipse x2 + 2y2 = 4 at points P and Q then the locus of the point of intersection of tangents at P and Q is :- a)a parabola with latus rectum = 4
- b)a parabola with focus as (2, 3)
- c)an ellipse with eccentricity
- d)an ellipse with eccentricity greater than 1/2
Correct answer is option 'C,D'. Can you explain this answer?
If a variable tangent of circle x2 + y2 = 1 intersects the ellipse x2 + 2y2 = 4 at points P and Q then the locus of the point of intersection of tangents at P and Q is :
a)
a parabola with latus rectum = 4
b)
a parabola with focus as (2, 3)
c)
an ellipse with eccentricity
d)
an ellipse with eccentricity greater than 1/2
|
|
Yash Patel answered |
In the figure shown strings AB and BC have masses m and 2m respectively. Both are of same length l. Mass of each string is uniformly distributed on its length. The string is suspended vertically from the ceiling of a room. A small jerk wave pulse is given at the end 'C'. It goes up to upper end 'A' in time 't'. If m = 2 kg, l = 
, Then find the value of 't' in seconds
Correct answer is '2'. Can you explain this answer?
In the figure shown strings AB and BC have masses m and 2m respectively. Both are of same length l. Mass of each string is uniformly distributed on its length. The string is suspended vertically from the ceiling of a room. A small jerk wave pulse is given at the end 'C'. It goes up to upper end 'A' in time 't'. If m = 2 kg, l = , Then find the value of 't' in seconds
, Then find the value of 't' in seconds
|
Telecom Tuners answered |
A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process.
Given specific heat of ice = 2100 J/kg-K
Given specific heat of water = 4200 J/kg-K
Latent heat of fusion of ice = 3.3 × 105 J/kg
Q. The initial mass of the ice in the container is equal to - a)0.36 kg
- b)1.2 kg
- c)2.4 kg
- d)3.6 kg
Correct answer is option 'C'. Can you explain this answer?
A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process.
Given specific heat of ice = 2100 J/kg-K
Given specific heat of water = 4200 J/kg-K
Latent heat of fusion of ice = 3.3 × 105 J/kg
Given specific heat of ice = 2100 J/kg-K
Given specific heat of water = 4200 J/kg-K
Latent heat of fusion of ice = 3.3 × 105 J/kg
Q. The initial mass of the ice in the container is equal to
a)
0.36 kg
b)
1.2 kg
c)
2.4 kg
d)
3.6 kg
|
Tanuja Kapoor answered |
Sample of ice has been receiving heat at constant rate P from water. Its temperature has increased by 30°C in time 60 min.
P is the foot of the perpendicular dropped from origin O on the line of intersection of the planes x – 2y + 3z = 5 and 2x + 3y + z + 4 = 0 then- a)
- b)Equation of acute angle bisector of the two planes is x + 5y – 2z + 9 = 0
- c)point (1, 2, 3) lies in acute region of the two planes.
- d)Equation of the plane perpendicular to the given line and passing through (1,1,1) is 11x – 5y – 7z + 1 = 0
Correct answer is option 'A,B,C,D'. Can you explain this answer?
P is the foot of the perpendicular dropped from origin O on the line of intersection of the planes x – 2y + 3z = 5 and 2x + 3y + z + 4 = 0 then
a)
b)
Equation of acute angle bisector of the two planes is x + 5y – 2z + 9 = 0
c)
point (1, 2, 3) lies in acute region of the two planes.
d)
Equation of the plane perpendicular to the given line and passing through (1,1,1) is 11x – 5y – 7z + 1 = 0
|
|
Anaya Patel answered |
An ellipse has semi-major axes of lengths 2 and semi-minor axis of length 1. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both x-axis and y-axis.
Q. The locus of foci of the ellipse is - a)
- b)
- c)x2+y2
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
An ellipse has semi-major axes of lengths 2 and semi-minor axis of length 1. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both x-axis and y-axis.
Q. The locus of foci of the ellipse is
a)
b)
c)
x2+y2
d)
None of these
|
|
Anaya Patel answered |
When the temperature of a copper coin is raised by 80 oC, its diameter increases by 0.2%, - a)percentage rise in the area of a face is 0.4%
- b)percentage rise in the thickness is 0.4%
- c)percentage rise in the volume is 0.6%
- d)coefficient of linear expansion of copper is 0.25x10-4 / oC.
Correct answer is option 'A,C,D'. Can you explain this answer?
When the temperature of a copper coin is raised by 80 oC, its diameter increases by 0.2%,
a)
percentage rise in the area of a face is 0.4%
b)
percentage rise in the thickness is 0.4%
c)
percentage rise in the volume is 0.6%
d)
coefficient of linear expansion of copper is 0.25x10-4 / oC.
|
|
Tejas Verma answered |
A long capillary tube of mass 'π' gm, radius 2mm and negligible thickness, is partially immersed in a liquid of surface tension 0.1 N/m. Take angle of contact zero and neglect buoyant force of liquid. The force required to hold the tube vertically, will be - (g = 10 m/s2) 
- a)10.4 π mN
- b)10.8 π mN
- c)0.8 π mN
- d)4.8 π mN
Correct answer is option 'B'. Can you explain this answer?
A long capillary tube of mass 'π' gm, radius 2mm and negligible thickness, is partially immersed in a liquid of surface tension 0.1 N/m. Take angle of contact zero and neglect buoyant force of liquid. The force required to hold the tube vertically, will be - (g = 10 m/s2)
a)
10.4 π mN
b)
10.8 π mN
c)
0.8 π mN
d)
4.8 π mN
|
Pranavi Iyer answered |
The number of common tangents to the hyperbola xy = c2 and the circle x2 + y2 = 2c2 is- a)4
- b)2
- c)8
- d)infinite
Correct answer is option 'B'. Can you explain this answer?
The number of common tangents to the hyperbola xy = c2 and the circle x2 + y2 = 2c2 is
a)
4
b)
2
c)
8
d)
infinite
|
Maitri Das answered |
Common Tangents between Hyperbola and Circle
- Equations:
The given hyperbola has the equation xy = c^2 and the circle has the equation x^2 + y^2 = 2c^2.
- Number of Tangents:
To find the number of common tangents between these two curves, we analyze their relative positions.
- Relative Positions:
Since the hyperbola is rectangular, it has asymptotes x = y and x = -y. The circle has its center at the origin and radius c√2.
- Analysis:
1. When the hyperbola is inside the circle, there are 4 common tangents.
2. When the hyperbola is outside the circle, there are 2 common tangents.
3. When the hyperbola is tangent to the circle, there are 8 common tangents.
- Conclusion:
In this case, the hyperbola xy = c^2 is outside the circle x^2 + y^2 = 2c^2, which results in 2 common tangents between them. Therefore, the correct answer is option B.
- Equations:
The given hyperbola has the equation xy = c^2 and the circle has the equation x^2 + y^2 = 2c^2.
- Number of Tangents:
To find the number of common tangents between these two curves, we analyze their relative positions.
- Relative Positions:
Since the hyperbola is rectangular, it has asymptotes x = y and x = -y. The circle has its center at the origin and radius c√2.
- Analysis:
1. When the hyperbola is inside the circle, there are 4 common tangents.
2. When the hyperbola is outside the circle, there are 2 common tangents.
3. When the hyperbola is tangent to the circle, there are 8 common tangents.
- Conclusion:
In this case, the hyperbola xy = c^2 is outside the circle x^2 + y^2 = 2c^2, which results in 2 common tangents between them. Therefore, the correct answer is option B.
Which of the following statement(s) is (are) correct for the thiosulphate salt ? - a)Thiosulphate ions reduce Fe3+ to Fe2+ in aqueous solution.
- b)It can be prepared by oxidation of polysulphides.
- c)The reaction of thiosulphate with iodine is the basis of iodometric titration in analytical chemistry.
- d)Chlorine oxidizes thiosulphate and this has led to the use of thiosulphate to remove excess chlorine in the bleaching industries.
Correct answer is option 'A,B,C,D'. Can you explain this answer?
Which of the following statement(s) is (are) correct for the thiosulphate salt ?
a)
Thiosulphate ions reduce Fe3+ to Fe2+ in aqueous solution.
b)
It can be prepared by oxidation of polysulphides.
c)
The reaction of thiosulphate with iodine is the basis of iodometric titration in analytical chemistry.
d)
Chlorine oxidizes thiosulphate and this has led to the use of thiosulphate to remove excess chlorine in the bleaching industries.
|
Prateek Nair answered |
An isosceles ΔABC is inscribed in the circle x2 + y2 = 16. From points A, B and C three ordinates are drawn
which cut the ellipse 
at the points P, Q and R respectively such that A and P are on the
opposite sides, Q and B are on the same side and C and R are on the same side of the major axis. If ΔPQR
is right angled at P and θ is the smallest angle of the ΔABC, then find the value of 16tan2θ + 8 tan θ + 9.
Correct answer is '24'. Can you explain this answer?
An isosceles ΔABC is inscribed in the circle x2 + y2 = 16. From points A, B and C three ordinates are drawn
which cut the ellipse at the points P, Q and R respectively such that A and P are on the
opposite sides, Q and B are on the same side and C and R are on the same side of the major axis. If ΔPQR
is right angled at P and θ is the smallest angle of the ΔABC, then find the value of 16tan2θ + 8 tan θ + 9.
which cut the ellipse
at the points P, Q and R respectively such that A and P are on theopposite sides, Q and B are on the same side and C and R are on the same side of the major axis. If ΔPQR
is right angled at P and θ is the smallest angle of the ΔABC, then find the value of 16tan2θ + 8 tan θ + 9.
|
|
Anaya Patel answered |
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π2 = 10; g = 10 m/s2) Assume that the spring does not affect the collision. 
Amplitude of the SHM as being executed by block B-spring system, is -- a)2.5√10 cm
- b)10 cm
- c)3√10cm
- d)5√10cm
Correct answer is option 'A'. Can you explain this answer?
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π2 = 10; g = 10 m/s2) Assume that the spring does not affect the collision.
Amplitude of the SHM as being executed by block B-spring system, is -
a)
2.5√10 cm
b)
10 cm
c)
3√10cm
d)
5√10cm
|
Veda Institute answered |
Immediately after the collision, suppose velocities of the blocks are V1 and V1 as shown 1/2 vel. of approach = velocity of separation.
⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or 10 = V1 + 5V2 .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or 10 = V1 + 5V2 .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
If all the words formed from the letters of the word "HORROR" are arranged in the opposite order as they are in a dictionary, then the rank of the word "HORROR" is- a)57
- b)58
- c)56
- d)59
Correct answer is option 'B'. Can you explain this answer?
If all the words formed from the letters of the word "HORROR" are arranged in the opposite order as they are in a dictionary, then the rank of the word "HORROR" is
a)
57
b)
58
c)
56
d)
59
|
Simran Chopra answered |
"ABRACADABRA" are arranged in alphabetical order and coded as in a dictionary, then what will be the code for the word "ABRACADABRA"?
In the figure shown ADB & BEF are two fixed circular paths. A block of mass m enters in the tube ADB through point A with minimum velocity to reach point B. From there it moves on another circular path of radius R'. There it is just able to complete the circle.
- a)velocity at A must be
- b)velocity at A must be
- c)
- d)the normal reaction at point E is 6mg
Correct answer is option 'D'. Can you explain this answer?
In the figure shown ADB & BEF are two fixed circular paths. A block of mass m enters in the tube ADB through point A with minimum velocity to reach point B. From there it moves on another circular path of radius R'. There it is just able to complete the circle.
a)
velocity at A must be
b)
velocity at A must be
c)
d)
the normal reaction at point E is 6mg
|
|
Ananya Das answered |
For minimum velocity. at A ;
For looping the loop ;
For looping the loop ;
In the shown circuit involving a resistor of resistance R W, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for t0 = RC seconds and then key is pushed back to position 1 for t0 = RC seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible.
Q. The charge on capacitor at t = 2RC second is - a)CE
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
In the shown circuit involving a resistor of resistance R W, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for t0 = RC seconds and then key is pushed back to position 1 for t0 = RC seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible.
Q. The charge on capacitor at t = 2RC second is
a)
CE
b)
c)
d)
|
Kamna Science Academy answered |
For t = 0 to t0 = RC seconds, the circuit is of charging type. The charging equation for this time is
Therefore the charge on capacitor at time t0 = RC is q0
For t = RC to t = 2RC seconds, the circuit is of discharging type. The charge and current equation for this time are
and 
Hence charge at t = 2 RC and current at t = 1.5 RC are
and
Since the capacitor gets more charged up from t = 2RC to t = 3RC than in the interval t = 0 to t = RC, the graph representing the charge variation is as shown in figure
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