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The equation of state of a gas is given by 
where P, V, T are pressure, volume and temperature respectively, and a, b, c are constants. The dimensions of a and b are respectively- a)ML8T–2 and L3/2
- b)ML5T–2 and L3
- c)ML5T–2 and L6
- d)ML6T–2 and L3/2
Correct answer is option 'A'. Can you explain this answer?
The equation of state of a gas is given by where P, V, T are pressure, volume and temperature respectively, and a, b, c are constants. The dimensions of a and b are respectively
where P, V, T are pressure, volume and temperature respectively, and a, b, c are constants. The dimensions of a and b are respectivelya)
ML8T–2 and L3/2
b)
ML5T–2 and L3
c)
ML5T–2 and L6
d)
ML6T–2 and L3/2
|
Muskan answered |
Just solve the two brackets separately by equating them with 0,i.e.( P+ a/v^3) =0 from here u will get the dimensions of a and( V-b^2)=0 from here u will get the dimensions of b...
Tautomerism is exhibited by- a)(Me3CCO)3CH
- b)
- c)
- d)
Correct answer is option 'A,B,D'. Can you explain this answer?
Tautomerism is exhibited by
a)
(Me3CCO)3CH
b)
c)
d)
|
|
Anand Kumar answered |
Tautomerism is shown by only those compounds which have at least one resonating hydrogen (tautomeric hydrogen).
In options A,B and D all structures have resonating hydrogen.So, these compounds will show tautomerism
In options A,B and D all structures have resonating hydrogen.So, these compounds will show tautomerism
The ore chromite is- a)FeCr2O4
- b)CoCr2O3
- c)CrFe2O4
- d)FeCr2O3
Correct answer is option 'A'. Can you explain this answer?
The ore chromite is
a)
FeCr2O4
b)
CoCr2O3
c)
CrFe2O4
d)
FeCr2O3
|
|
Kiran Sarmah answered |
Chromite ore contains chromium, iron and oxygen. Its formula is FeCr2O4.
The best method for preparation of Me3CCN is- a)To react Me3COH with HCN
- b)To react Me3CBr with NaCN
- c)To react Me3CMgBr with ClCN
- d)To react Me3CLi with NH2CN
Correct answer is option 'C'. Can you explain this answer?
The best method for preparation of Me3CCN is
a)
To react Me3COH with HCN
b)
To react Me3CBr with NaCN
c)
To react Me3CMgBr with ClCN
d)
To react Me3CLi with NH2CN
|
Jaideep Mukherjee answered |
Preparation of Me3CCN
Reacting Me3CMgBr with ClCN is the best method for the preparation of Me3CCN. This reaction involves the use of the Grignard reagent, Me3CMgBr, which is a strong nucleophile and reacts with ClCN to form Me3CCN.
Steps Involved
The reaction proceeds through the following steps:
1. Formation of the Grignard reagent: Me3CMgBr is prepared by reacting Me3CBr with Mg in ether.
2. Reaction with ClCN: Me3CMgBr is then reacted with ClCN in dry ether to form Me3CCN.
3. Workup: The reaction mixture is then treated with dilute acid to hydrolyze any unreacted Grignard reagent and to protonate the product, Me3CCN.
Mechanism
The reaction mechanism involves the attack of the nucleophilic carbon atom of the Grignard reagent on the electrophilic carbon atom of ClCN. This leads to the formation of an intermediate, which then undergoes hydrolysis to form the final product, Me3CCN.
Advantages of this method
1. High yields: This method gives high yields of Me3CCN.
2. Mild conditions: The reaction conditions are relatively mild, and the reaction can be carried out at room temperature.
3. Availability of reagents: The reagents used in the reaction, Me3CBr, Mg, and ClCN, are readily available and inexpensive.
4. Versatility: The Grignard reagent, Me3CMgBr, can be used for the preparation of a variety of other organic compounds.
Thus, the reaction of Me3CMgBr with ClCN is the best method for the preparation of Me3CCN.
Reacting Me3CMgBr with ClCN is the best method for the preparation of Me3CCN. This reaction involves the use of the Grignard reagent, Me3CMgBr, which is a strong nucleophile and reacts with ClCN to form Me3CCN.
Steps Involved
The reaction proceeds through the following steps:
1. Formation of the Grignard reagent: Me3CMgBr is prepared by reacting Me3CBr with Mg in ether.
2. Reaction with ClCN: Me3CMgBr is then reacted with ClCN in dry ether to form Me3CCN.
3. Workup: The reaction mixture is then treated with dilute acid to hydrolyze any unreacted Grignard reagent and to protonate the product, Me3CCN.
Mechanism
The reaction mechanism involves the attack of the nucleophilic carbon atom of the Grignard reagent on the electrophilic carbon atom of ClCN. This leads to the formation of an intermediate, which then undergoes hydrolysis to form the final product, Me3CCN.
Advantages of this method
1. High yields: This method gives high yields of Me3CCN.
2. Mild conditions: The reaction conditions are relatively mild, and the reaction can be carried out at room temperature.
3. Availability of reagents: The reagents used in the reaction, Me3CBr, Mg, and ClCN, are readily available and inexpensive.
4. Versatility: The Grignard reagent, Me3CMgBr, can be used for the preparation of a variety of other organic compounds.
Thus, the reaction of Me3CMgBr with ClCN is the best method for the preparation of Me3CCN.
Acid catalysed hydrolysis of ethyl acetate follows a pseudo-first order kinetics with respect to ester. If the reaction is carried out with large excess of ester, the order with respect to ester will be- a)1.5
- b)0
- c)2
- d)1
Correct answer is option 'B'. Can you explain this answer?
Acid catalysed hydrolysis of ethyl acetate follows a pseudo-first order kinetics with respect to ester. If the reaction is carried out with large excess of ester, the order with respect to ester will be
a)
1.5
b)
0
c)
2
d)
1
|
|
Ujjwal Mishra answered |
Because at excess of water the concentration of ether is approx negligible and the reaction does not depends on the concentration of ether hence the reaction follow 0 order reaction
A particle of mass M and charge q is at rest at the midpoint between two other fixed similar charges each of magnitude Q placed a distance 2d apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount x (x<< d) along the line joining the two charges and is left to itself. It will now oscillate about the mean position with a time period (ε0 = permittivity of free space)
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A particle of mass M and charge q is at rest at the midpoint between two other fixed similar charges each of magnitude Q placed a distance 2d apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount x (x<< d) along the line joining the two charges and is left to itself. It will now oscillate about the mean position with a time period (ε0 = permittivity of free space)
a)
b)
c)
d)
|
Chirag Verma answered |
Restoring force on displacement of x,
A current of 1 A is flowing along positive x-axis through a straight wire of length 0.5 m placed in a region of a magnetic field given by B =(2ˆi+ 4ˆj) T. The magnitude and the direction of the force experienced by the wire respectively are- a)√18N , along positive z-axis
- b)√20N , along positive x-axis
- c)2N, along positive z-axis
- d) 4N, along positive y-axis
Correct answer is option 'C'. Can you explain this answer?
A current of 1 A is flowing along positive x-axis through a straight wire of length 0.5 m placed in a region of a magnetic field given by B =(2ˆi+ 4ˆj) T. The magnitude and the direction of the force experienced by the wire respectively are
a)
√18N , along positive z-axis
b)
√20N , along positive x-axis
c)
2N, along positive z-axis
d)
4N, along positive y-axis
|
|
Chirag Verma answered |
The equation 2x2 + 5xy – 12y2 = 0 represents a- a)circle
- b)pair of non-perpendicular intersecting straight lines
- c)pair of perpendicular straight lines
- d)hyperbola
Correct answer is option 'B'. Can you explain this answer?
The equation 2x2 + 5xy – 12y2 = 0 represents a
a)
circle
b)
pair of non-perpendicular intersecting straight lines
c)
pair of perpendicular straight lines
d)
hyperbola
|
Rajesh Karsanbhi answered |
2x2-12y2+5xy=o
(x+5/4y)2-25/4y2-6y2=0
{x+(5/4)y}2=(49/4)y2
x+5/4y=7/2y. or x+5/4y=-7/2y
this not perpenticuler
(x+5/4y)2-25/4y2-6y2=0
{x+(5/4)y}2=(49/4)y2
x+5/4y=7/2y. or x+5/4y=-7/2y
this not perpenticuler
Lines x + y = 1 and 3y = x + 3 intersect the ellipse x2 + 9y2 = 9 at the points P,Q,R. The are of the triangle PQR is- a)36/5
- b) 18/5
- c) 9/5
- d)1/5
Correct answer is option 'B'. Can you explain this answer?
Lines x + y = 1 and 3y = x + 3 intersect the ellipse x2 + 9y2 = 9 at the points P,Q,R. The are of the triangle PQR is
a)
36/5
b)
18/5
c)
9/5
d)
1/5
|
Nikhil Banerjee answered |
Problem: Find the area of the triangle PQR formed by the intersection of lines x y = 1 and 3y = x-3 with the ellipse x^2/9 + y^2 = 1.
Solution:
To solve the problem, we need to find the points P, Q, and R of intersection between the given lines and the ellipse.
Step 1: Find the points of intersection of the ellipse and the line x y = 1.
Substituting y = 1/x in the equation of the ellipse, we get:
x^2/9 + (1/x)^2 = 1
Multiplying both sides by x^2, we get:
x^4/9 + 1 = x^2
Bringing all terms to one side, we get:
x^4 - 9x^2 + 9 = 0
This is a quadratic equation in x^2. Solving for x^2 using the quadratic formula, we get:
x^2 = (9 ± 3√5)/2
Since x^2 is positive, we take the positive root and get:
x^2 = (9 + 3√5)/2
Taking the square root, we get:
x = √[(9 + 3√5)/2]
Substituting this value of x in the equation x y = 1, we get:
y = 1/x = √[2/(9 + 3√5)]
So, the point of intersection of the ellipse and the line x y = 1 is:
P = (√[(9 + 3√5)/2], √[2/(9 + 3√5)])
Similarly, we can find the point of intersection of the ellipse and the line 3y = x-3.
Step 2: Find the area of the triangle PQR.
Let Q be the point of intersection of the ellipse and the line 3y = x-3.
Substituting y = (x-3)/3 in the equation of the ellipse, we get:
x^2/9 + ((x-3)/3)^2 = 1
Simplifying, we get:
4x^2 - 36x + 36 = 0
Solving for x using the quadratic formula, we get:
x = 3/2 ± √(5)/2
Since the lines x y = 1 and 3y = x-3 intersect at the point (3, 0), we can easily see that the point of intersection of the ellipse and the line 3y = x-3 is:
R = (3/2 - √(5)/2, 3/2)
Now, we can find the lengths of the sides of the triangle PQR using the distance formula.
PR = √[(√[(9 + 3√5)/2] - 3/2 + √(5)/2)^2 + (√[2/(9 + 3√5)] - 3/2)^2]
QR = √[(3/2 + √(5)/2 - (3/2 - √(5)/2))^2 + (3/2 - 0)^2]
Solution:
To solve the problem, we need to find the points P, Q, and R of intersection between the given lines and the ellipse.
Step 1: Find the points of intersection of the ellipse and the line x y = 1.
Substituting y = 1/x in the equation of the ellipse, we get:
x^2/9 + (1/x)^2 = 1
Multiplying both sides by x^2, we get:
x^4/9 + 1 = x^2
Bringing all terms to one side, we get:
x^4 - 9x^2 + 9 = 0
This is a quadratic equation in x^2. Solving for x^2 using the quadratic formula, we get:
x^2 = (9 ± 3√5)/2
Since x^2 is positive, we take the positive root and get:
x^2 = (9 + 3√5)/2
Taking the square root, we get:
x = √[(9 + 3√5)/2]
Substituting this value of x in the equation x y = 1, we get:
y = 1/x = √[2/(9 + 3√5)]
So, the point of intersection of the ellipse and the line x y = 1 is:
P = (√[(9 + 3√5)/2], √[2/(9 + 3√5)])
Similarly, we can find the point of intersection of the ellipse and the line 3y = x-3.
Step 2: Find the area of the triangle PQR.
Let Q be the point of intersection of the ellipse and the line 3y = x-3.
Substituting y = (x-3)/3 in the equation of the ellipse, we get:
x^2/9 + ((x-3)/3)^2 = 1
Simplifying, we get:
4x^2 - 36x + 36 = 0
Solving for x using the quadratic formula, we get:
x = 3/2 ± √(5)/2
Since the lines x y = 1 and 3y = x-3 intersect at the point (3, 0), we can easily see that the point of intersection of the ellipse and the line 3y = x-3 is:
R = (3/2 - √(5)/2, 3/2)
Now, we can find the lengths of the sides of the triangle PQR using the distance formula.
PR = √[(√[(9 + 3√5)/2] - 3/2 + √(5)/2)^2 + (√[2/(9 + 3√5)] - 3/2)^2]
QR = √[(3/2 + √(5)/2 - (3/2 - √(5)/2))^2 + (3/2 - 0)^2]
The de Broglie wavelength of an electron (mass = 1 × 10–30 kg, charge = 1.6 × 10-19 C) with a kinetic energy of 200 eV is (Planck’s constant = 6.6 × 10–34 J s)- a)9.60 × 10–11 m
- b)8.25 × 10–11 m
- c)6.25 × 10–11 m
- d)5.00 × 10–11 m
Correct answer is option 'B'. Can you explain this answer?
The de Broglie wavelength of an electron (mass = 1 × 10–30 kg, charge = 1.6 × 10-19 C) with a kinetic energy of 200 eV is (Planck’s constant = 6.6 × 10–34 J s)
a)
9.60 × 10–11 m
b)
8.25 × 10–11 m
c)
6.25 × 10–11 m
d)
5.00 × 10–11 m
|
Priyal Khandelwal answered |
H/p= lamda direct formula based
Correct statement(s) in cases of n-butanol and t-butanol is (are)- a)Both are having equal solubility in water
- b)t-butanol is more soluble in water than n-butanol
- c)Boiling point of t-butanol is lower than n-butanol
- d)Boiling point of n-butanol is lower than t-butanol
Correct answer is option 'B,C'. Can you explain this answer?
Correct statement(s) in cases of n-butanol and t-butanol is (are)
a)
Both are having equal solubility in water
b)
t-butanol is more soluble in water than n-butanol
c)
Boiling point of t-butanol is lower than n-butanol
d)
Boiling point of n-butanol is lower than t-butanol
|
|
Anand Kumar answered |
The increasing order of O-N-O bond angle in the species NO2+ , NO2and NO2– is- a)NO2+<NO2<NO2–
- b)NO2<NO2–<NO2+
- c)NO2+<NO2–<NO2
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
The increasing order of O-N-O bond angle in the species NO2+ , NO2and NO2– is
a)
NO2+<NO2<NO2–
b)
NO2<NO2–<NO2+
c)
NO2+<NO2–<NO2
d)
none of these
|
Rohit Jain answered |
No option is correct correct ans : NO2+ > NO2 > NO2–
Equivalent conductivity at infinite dilution for sodium-potassium oxalate ((COO–)2Na+K+) will be [given, molar conductivities of oxalate, K+ and Na+ ions at infinite dilution are 148.2, 50.1, 73.5 S cm2mol–1, respectively]- a)271.8 S cm2 eq–1
- b)67.95 S cm2 eq–1
- c)543.6 S cm2 eq–1
- d)135.9 S cm2 eq–1
Correct answer is option 'D'. Can you explain this answer?
Equivalent conductivity at infinite dilution for sodium-potassium oxalate ((COO–)2Na+K+) will be [given, molar conductivities of oxalate, K+ and Na+ ions at infinite dilution are 148.2, 50.1, 73.5 S cm2mol–1, respectively]
a)
271.8 S cm2 eq–1
b)
67.95 S cm2 eq–1
c)
543.6 S cm2 eq–1
d)
135.9 S cm2 eq–1
|
Gaurav Kumar answered |
An objective type test paper has 5 questions. Out of these 5 questions, 3 questions have four options each (A, B, C, D) with one option being the correct answer. The other 2 questions have two options each, namely True and False. A candidate randomly ticks the options. Then the probability that he/she will tick the correct option in at least four questions, is- a) 5/32
- b) 3/128
- c) 3/256
- d)3/64
Correct answer is option 'D'. Can you explain this answer?
An objective type test paper has 5 questions. Out of these 5 questions, 3 questions have four options each (A, B, C, D) with one option being the correct answer. The other 2 questions have two options each, namely True and False. A candidate randomly ticks the options. Then the probability that he/she will tick the correct option in at least four questions, is
a)
5/32
b)
3/128
c)
3/256
d)
3/64
|
Avi Verma answered |
n(S) = 43. 22, n(e) = (3C1 .3 + 2C1. 1) + 1
Let z1 = 2 + 3i and z2 = 3 + 4i be two points on the complex plane. Then the set of complex numbers z satisfying |z – z1|2 + |z – z2|2 = |z1 – z2|2 represents- a)a straight line
- b)a point
- c)a circle
- d)a pair of straight lines
Correct answer is option 'C'. Can you explain this answer?
Let z1 = 2 + 3i and z2 = 3 + 4i be two points on the complex plane. Then the set of complex numbers z satisfying |z – z1|2 + |z – z2|2 = |z1 – z2|2 represents
a)
a straight line
b)
a point
c)
a circle
d)
a pair of straight lines
|
Gayatri Bajaj answered |
Clearly the locus of Z is a circle with Z1 & Z2 as end point of diameter.
There are two coins, one unbiased with probability 1/2 of getting heads and the other one is biased with probability 3/4 of getting heads. A coin is selected at random and tossed. It shows heads up. Then the probability that the unbiased coin was selected is- a)2/3
- b)3/5
- c)1/2
- d)2/5
Correct answer is option 'D'. Can you explain this answer?
There are two coins, one unbiased with probability 1/2 of getting heads and the other one is biased with probability 3/4 of getting heads. A coin is selected at random and tossed. It shows heads up. Then the probability that the unbiased coin was selected is
a)
2/3
b)
3/5
c)
1/2
d)
2/5
|
|
Bijoy Sen answered |
H → Event of head showing up
B → Event of biased coin chosen
UB → Event of unbiased coin chosen
B → Event of biased coin chosen
UB → Event of unbiased coin chosen
An electric cell of e.m.f. E is connected across a copper wire of diameter d and length l. The drift velocity of electrons in the wire is vd. If the length of the wire is changed to 2l, the new drift velocity of electrons in the copper wire will be- a)νd
- b)2νd
- c)νd/2
- d)νd/4
Correct answer is option 'C'. Can you explain this answer?
An electric cell of e.m.f. E is connected across a copper wire of diameter d and length l. The drift velocity of electrons in the wire is vd. If the length of the wire is changed to 2l, the new drift velocity of electrons in the copper wire will be
a)
νd
b)
2νd
c)
νd/2
d)
νd/4
|
Rhea Joshi answered |
The specific heat c of a solid at low temperature shows temperature dependence according to the relation c = DT3 where D is a constant and T is the temperature in kelvin. A piece of this solid of mass m kg is taken and its temperature is raised from 20 K to 30 K. The amount of the heat required in the process in energy units is- a)5 × 104 Dm
- b)(33/4) × 104 Dm
- c)(65/4) × 104 Dm
- d)(5/4) × 104 Dm
Correct answer is option 'C'. Can you explain this answer?
The specific heat c of a solid at low temperature shows temperature dependence according to the relation c = DT3 where D is a constant and T is the temperature in kelvin. A piece of this solid of mass m kg is taken and its temperature is raised from 20 K to 30 K. The amount of the heat required in the process in energy units is
a)
5 × 104 Dm
b)
(33/4) × 104 Dm
c)
(65/4) × 104 Dm
d)
(5/4) × 104 Dm
|
Amit Kumar answered |
A particle of mass M and charge q, initially at rest, is accelerated by a uniform electric field E through a distance D and is then allowed to approach a fixed static charge Q of the same sign. The distance of the closest approach of the charge q will then be- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A particle of mass M and charge q, initially at rest, is accelerated by a uniform electric field E through a distance D and is then allowed to approach a fixed static charge Q of the same sign. The distance of the closest approach of the charge q will then be
a)
b)
c)
d)
|
|
Ram Mohith answered |
A van der Waals gas may behave ideally when- a)The volume is very low
- b)The temperature is very high
- c)The pressure is very low
- d)The temperature, pressure and volume all are very high
Correct answer is option 'C'. Can you explain this answer?
A van der Waals gas may behave ideally when
a)
The volume is very low
b)
The temperature is very high
c)
The pressure is very low
d)
The temperature, pressure and volume all are very high
|
|
Anjali Sharma answered |
A van der waals gas may behave ideally when pressure is very low as compressibility factor (Z) approaches 1. At high temperature Z > 1.
The optically active molecule is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The optically active molecule is
a)
b)
c)
d)
|
|
Khaja Moinuddin answered |
All the carbon atoms is all options are chiral but in options A,B,D there is a plane of symmetry which makes them optically inactive..so the answer is C..it does not have plane of symmetry
For BCl3, AlCl3 and GaCl3 the increasing order of ionic character is- a)BCl3<AlCl3<GaCl3
- b)GaCl3<AlCl3<BCl3
- c)BCl3<GaCl3<AlCl3
- d)AlCl3<BCl3<GaCl3
Correct answer is option 'C'. Can you explain this answer?
For BCl3, AlCl3 and GaCl3 the increasing order of ionic character is
a)
BCl3<AlCl3<GaCl3
b)
GaCl3<AlCl3<BCl3
c)
BCl3<GaCl3<AlCl3
d)
AlCl3<BCl3<GaCl3
|
Jyoti Patel answered |
Ionic character is inversely proportional to polarising power of cation.
BCl3<GaCl3<AlCl3
The number of atoms of a radioactive substance of half-life T is N0 at t = 0. The time necessary to decay from N0/2 atoms to N0/10 atoms will be- a)5T/2
- b)Tln5
- c)Tln(5/2)
- d)T ln5/ln2
Correct answer is option 'C'. Can you explain this answer?
The number of atoms of a radioactive substance of half-life T is N0 at t = 0. The time necessary to decay from N0/2 atoms to N0/10 atoms will be
a)
5T/2
b)
Tln5
c)
Tln(5/2)
d)
T ln5/ln2
|
Upasana Sengupta answered |
Explanation:
- Half-life of a radioactive substance:
- The half-life of a radioactive substance is the time required for half of the radioactive atoms in a sample to decay.
- Given:
- Number of atoms at t = 0: N0
- Time to decay from N0/2 atoms to N0/10 atoms
- Time taken to decay from N0 to N0/2:
- Since the half-life is T, it takes T time to decay from N0 to N0/2.
- Time taken to decay from N0/2 to N0/10:
- To decay from N0/2 to N0/10, we need to find the time taken to decay one half-life further.
- When N0/2 atoms decay to N0/4 atoms, one half-life has passed.
- When N0/4 atoms decay to N0/8 atoms, another half-life has passed.
- When N0/8 atoms decay to N0/10 atoms, it is half of the remaining atoms, which means half-life has passed again.
- Calculating the total time:
- To decay from N0/2 to N0/10, a total of 3 half-lives have passed (N0/2 -> N0/4 -> N0/8 -> N0/10).
- The total time taken will be 3 times the half-life, which is 3T.
- Conclusion:
- The time necessary to decay from N0/2 atoms to N0/10 atoms is 3T.
- Therefore, the correct answer is option 'C' - Tln(5/2).
- Half-life of a radioactive substance:
- The half-life of a radioactive substance is the time required for half of the radioactive atoms in a sample to decay.
- Given:
- Number of atoms at t = 0: N0
- Time to decay from N0/2 atoms to N0/10 atoms
- Time taken to decay from N0 to N0/2:
- Since the half-life is T, it takes T time to decay from N0 to N0/2.
- Time taken to decay from N0/2 to N0/10:
- To decay from N0/2 to N0/10, we need to find the time taken to decay one half-life further.
- When N0/2 atoms decay to N0/4 atoms, one half-life has passed.
- When N0/4 atoms decay to N0/8 atoms, another half-life has passed.
- When N0/8 atoms decay to N0/10 atoms, it is half of the remaining atoms, which means half-life has passed again.
- Calculating the total time:
- To decay from N0/2 to N0/10, a total of 3 half-lives have passed (N0/2 -> N0/4 -> N0/8 -> N0/10).
- The total time taken will be 3 times the half-life, which is 3T.
- Conclusion:
- The time necessary to decay from N0/2 atoms to N0/10 atoms is 3T.
- Therefore, the correct answer is option 'C' - Tln(5/2).
Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation is- a)2ma2
- b)4ma2
- c)2m(a2 + b2)
- d)2m(a2 – b2)
Correct answer is option 'A'. Can you explain this answer?
Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation is
a)
2ma2
b)
4ma2
c)
2m(a2 + b2)
d)
2m(a2 – b2)
|
|
Raghavendra Ghoshal answered |
Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where E is internal energy of the system)- a)–ΔP/P
- b)Zero
- c)–VΔP
- d)–ΔE
Correct answer is option 'B'. Can you explain this answer?
Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where E is internal energy of the system)
a)
–ΔP/P
b)
Zero
c)
–VΔP
d)
–ΔE
|
|
Ram Mohith answered |
The work done by gas in any process is given by the integral of pressure P and small change in volume dv. Or simply the small work, dw done by the gas is given as
dw = P dv
In a constant volume process there is no change in volume so dv = 0 and hence work done by gas is zero.
dw = P dv
In a constant volume process there is no change in volume so dv = 0 and hence work done by gas is zero.
A particle is moving with a uniform speed v in a circular path of radius r with the centre at O. When the particle moves from a point P to Q on the circle such that ∠POQ = θ, then the magnitude of the change in velocity is- a)2v sin (2θ)
- b)zero
- c)2v sin (θ/2)
- d)2v cos (θ/2)
Correct answer is option 'C'. Can you explain this answer?
A particle is moving with a uniform speed v in a circular path of radius r with the centre at O. When the particle moves from a point P to Q on the circle such that ∠POQ = θ, then the magnitude of the change in velocity is
a)
2v sin (2θ)
b)
zero
c)
2v sin (θ/2)
d)
2v cos (θ/2)
|
|
Amit Kumar answered |
For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R is- a)Reflexive but not Symmetric
- b)Symmetric but not transitive
- c)Transitive but not Reflexive
- d)an Equivalence relation
Correct answer is option 'D'. Can you explain this answer?
For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R is
a)
Reflexive but not Symmetric
b)
Symmetric but not transitive
c)
Transitive but not Reflexive
d)
an Equivalence relation
|
|
Upasana Sengupta answered |
sin2a + cos2b = 1
Reflexive : sin2a + cos2a = 1
⇒ aRa
sin2a + cos2b = 1,
1 – cos2a + 1 – sin2b
= 1
sin2b + cos2a = 1
⇒ bRa
Hence symmetric Let aRb, bRc
sin2a + cos2b = 1 ............. (1)
sin2b + cos2c = 1 ............... (2)
(1) + (2)
sin2a + cos2c = 1
Hence transitive therefore equivalence relation.
Reflexive : sin2a + cos2a = 1
⇒ aRa
sin2a + cos2b = 1,
1 – cos2a + 1 – sin2b
= 1
sin2b + cos2a = 1
⇒ bRa
Hence symmetric Let aRb, bRc
sin2a + cos2b = 1 ............. (1)
sin2b + cos2c = 1 ............... (2)
(1) + (2)
sin2a + cos2c = 1
Hence transitive therefore equivalence relation.
A family of curves is such that the length intercepted on the y-axis between the origin and the tangent at a point is three times the ordinate of the point of contact. The family of curves is- a)xy = c, c is a constant
- b)xy2 = c, c is a constant
- c)x2y = c, c is a constant
- d)x2y2 = c, c is a constant
Correct answer is option 'C'. Can you explain this answer?
A family of curves is such that the length intercepted on the y-axis between the origin and the tangent at a point is three times the ordinate of the point of contact. The family of curves is
a)
xy = c, c is a constant
b)
xy2 = c, c is a constant
c)
x2y = c, c is a constant
d)
x2y2 = c, c is a constant
|
|
Ram Mohith answered |
Each of a and b can take values 1 or 2 with equal probability. The probability that the equation ax2 + bx + 1 = 0 has real roots, is equal to- a)1/2
- b)1/4
- c)1/8
- d)1/16
Correct answer is option 'B'. Can you explain this answer?
Each of a and b can take values 1 or 2 with equal probability. The probability that the equation ax2 + bx + 1 = 0 has real roots, is equal to
a)
1/2
b)
1/4
c)
1/8
d)
1/16
|
Vandana Bajaj answered |
Solution:
The given equation is ax² + bx + 1 = 0.
For the equation to have real roots, the discriminant b² - 4ac should be greater than or equal to zero.
Substituting the values of a and b, we get
b² - 4ac = b² - 4a
Since a and b can take values 1 or 2 with equal probability, the probability of b² - 4a being greater than or equal to zero is the same as the probability of b² - 4a being less than zero.
Therefore, P(b² - 4a ≥ 0) = P(b² - 4a < 0)="" />
Now, we need to find the probability that the discriminant is greater than or equal to zero, i.e., P(b² - 4ac ≥ 0).
Case 1: a = 1
If a = 1, then the discriminant becomes b² - 4c.
For the discriminant to be greater than or equal to zero, we need b² ≥ 4c.
Since b can take values 1 or 2 with equal probability, the probability of b² ≥ 4c is the same as the probability of b² < />
Therefore, P(b² - 4c ≥ 0) = P(b² - 4c < 0)="" />
Case 2: a = 2
If a = 2, then the discriminant becomes b² - 8c.
For the discriminant to be greater than or equal to zero, we need b² ≥ 8c.
Since b can take values 1 or 2 with equal probability, the probability of b² ≥ 8c is the same as the probability of b² < />
Therefore, P(b² - 8c ≥ 0) = P(b² - 8c < 0)="" />
The total probability of the equation having real roots is the sum of the probabilities from both cases:
P(ax² + bx + 1 = 0 has real roots) = P(a = 1) × P(b² - 4c ≥ 0) + P(a = 2) × P(b² - 8c ≥ 0)
= (1/2) × (1/2) + (1/2) × (1/2) = 1/4
Therefore, the correct option is (B) 1/4.
The given equation is ax² + bx + 1 = 0.
For the equation to have real roots, the discriminant b² - 4ac should be greater than or equal to zero.
Substituting the values of a and b, we get
b² - 4ac = b² - 4a
Since a and b can take values 1 or 2 with equal probability, the probability of b² - 4a being greater than or equal to zero is the same as the probability of b² - 4a being less than zero.
Therefore, P(b² - 4a ≥ 0) = P(b² - 4a < 0)="" />
Now, we need to find the probability that the discriminant is greater than or equal to zero, i.e., P(b² - 4ac ≥ 0).
Case 1: a = 1
If a = 1, then the discriminant becomes b² - 4c.
For the discriminant to be greater than or equal to zero, we need b² ≥ 4c.
Since b can take values 1 or 2 with equal probability, the probability of b² ≥ 4c is the same as the probability of b² < />
Therefore, P(b² - 4c ≥ 0) = P(b² - 4c < 0)="" />
Case 2: a = 2
If a = 2, then the discriminant becomes b² - 8c.
For the discriminant to be greater than or equal to zero, we need b² ≥ 8c.
Since b can take values 1 or 2 with equal probability, the probability of b² ≥ 8c is the same as the probability of b² < />
Therefore, P(b² - 8c ≥ 0) = P(b² - 8c < 0)="" />
The total probability of the equation having real roots is the sum of the probabilities from both cases:
P(ax² + bx + 1 = 0 has real roots) = P(a = 1) × P(b² - 4c ≥ 0) + P(a = 2) × P(b² - 8c ≥ 0)
= (1/2) × (1/2) + (1/2) × (1/2) = 1/4
Therefore, the correct option is (B) 1/4.
A mass M at rest is broken into two pieces having masses m and (M-m). The two masses are then separated by a distance r. The gravitational force between them will be the maximum when the ratio of the masses [m:(M-m)] of the two parts is- a)1 : 1
- b)1 : 2
- c)1: 3
- d)1 : 4
Correct answer is option 'A'. Can you explain this answer?
A mass M at rest is broken into two pieces having masses m and (M-m). The two masses are then separated by a distance r. The gravitational force between them will be the maximum when the ratio of the masses [m:(M-m)] of the two parts is
a)
1 : 1
b)
1 : 2
c)
1: 3
d)
1 : 4
|
|
Magnus Gaaruna answered |
2-Methylpropane on monochlorination under photochemical condition give- a)2-Chloro-2-methylpropane as major product
- b)(1:1) Mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane
- c)1-Chloro-2-methylpropane as a major product
- d)(1:9) Mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane
Correct answer is option 'C'. Can you explain this answer?
2-Methylpropane on monochlorination under photochemical condition give
a)
2-Chloro-2-methylpropane as major product
b)
(1:1) Mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane
c)
1-Chloro-2-methylpropane as a major product
d)
(1:9) Mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane
|
|
Janani Menon answered |
Ratio of A: B is 5:9
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