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The coefficient of linear expansion for a certain metal varies with absolute temperature as α = aT. If L_{o} = 1 m is the initial length of the metal and the temperature of metal is changed from 27^{o}C to 127^{o}C, then final length is (where a = 2 × 10^{5} K^{−2})
A billiard player hits a billiard ball of mass m and radius r lying at rest on the billiard table. When the cue hits the ball horizontally with force F at a height (r + h) from the table top the ball rolls without slipping. h is given as
Assuming zaxis to be coming out of the plane of the paper
A conducting rod is rotated by means of strings in a uniform magnetic field with constant angular velocity as shown in the figure. Potential of points A, B and C are V_{A}, V_{B} and V_{C} respectively. Then
Points A and C will rotate on the same circle. Thus V_{A} = V_{C} and V_{A} > V_{B}.
Let us split the circuit and then rearrange it as shown:
Since this represents a balanced wheatstone bridge, current in CD branch is zero.
A Π s haped wire frame with a light, smooth, sliding wire connected to its open end (width of open end is l) is dipped in a soap solution and then raised, such that a soap film forms between the slider and the closed end of the wire frame. If the wire frame is now placed in a vertical plane and a block of mass m is hung from the slider then for mass m to remain in equilibrium the surface tension of the soap film is
Force due to surface tension,
F = 2T × l (since there are two free surfaces)
For equilibrium F = mg
⇒ T = mg/2l
An artificial satellite of mass m is moving in a circular orbit at a height equal to the radius R of the Earth. Suddenly due to internal explosion the satellite breaks into two parts of equal masses. One part of the satellite stops just after the explosion and then falls to the surface of the Earth. The increase in the mechanical energy of the system (satellite + Earth) due to explosion will be (take acceleration due to gravity on the surface of Earth as g)
Conserving momentum during the explosion
Increase in mechanical energy is ΔE = ΔK + ΔU = ΔK + 0
In position A kinetic energy of a particle is 60 J and potential energy is −20 J. In position B, kinetic energy is 100 J and potential energy is 40 J. Then in moving the particle from A to B
Work done by conservative forces
= U_{i} − U_{f} = −20 − 40 = −60 J
Work done by external forces
= E_{f} – E_{i} = (K_{f} + U_{f}) – (K_{i} + U_{i}) = 40 – 140 = −100 J
Work done by all the forces
= K_{f} – K_{i} = 100 – 60 = 40 J
A metal rod is fixed in horizontal position and a force of magnitude F is applied as shown. If R_{A} = force by wall A and R_{B} = force by wall B, then
F. B. D. of the two parts of the rod are as shown.
Also, increase in length of 1 = decrease in length of 2
Two waves travelling in opposite directions produce a standing wave. The individual wave functions are given by y_{1} = 4 sin (3x – 2t) cm and y_{2} = 4 sin (3x + 2t) cm, where x and y are in centimeter. Now, select the correct statement(s) from the following.
y = y_{1} + y_{2} = (8 sin 3x. cos 2t) = A_{x}cos 2t
Where A_{x} = 8 sin 3x
∴ Nodes are formed where 3x = 0, π, 2π, ….. etc.
A point source S is placed anywhere in between two converging mirrors having focal lengths f and 2f, respectively as shown. The value of d for which only single image may be formed is /are
Only one image will be formed if the object is either at the common focus or common centre of curvature of both the mirrors. The ray diagrams are as shown.
In the circuit shown, initially keys K_{1}, K_{2} and K_{3} are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K_{1}, K_{2} and K_{3}.
Event 1: Keys K_{1} and K_{2} are closed and K_{3} is left open.
Event 2: After a long time K_{2} is opened and K_{3} is closed.
Event 3: After a long time K_{1} is opened and K_{2} is closed.
Q.
The time constants for charging of capacitor (completion of Event 1) and rise of current through the inductor (completion of Event 2) respectively are
Equivalent circuit after completion of Event 1 is
Equivalent circuit after completion of Event 2 is
In the circuit shown, initially keys K_{1}, K_{2} and K_{3} are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K_{1}, K_{2} and K_{3}.
Event 1: Keys K_{1} and K_{2} are closed and K_{3} is left open.
Event 2: After a long time K_{2} is opened and K_{3} is closed.
Event 3: After a long time K_{1} is opened and K_{2} is closed.
Q.
Find the maximum charge that can come on the capacitor after the completion of all the three events
Steady state charge on capacitor is
Steady state current (I) through the inductor is
Thus the capacitor again starts charging.
Charge on the capacitor will reach maximum value when current through the inductor becomes zero.
From conservation of energy between initial state (when i = 5 A, q = 40 μC) and final state (when i = 0, q = Q_{max}) we get
In the circuit shown, initially keys K_{1}, K_{2} and K_{3} are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K_{1}, K_{2} and K_{3}.
Event 1: Keys K_{1} and K_{2} are closed and K_{3} is left open.
Event 2: After a long time K_{2} is opened and K_{3} is closed.
Event 3: After a long time K_{1} is opened and K_{2} is closed.
Q.
Find the maximum current that can pass through the inductor after the completion of all the three events
After completion of Event 3
The capacitor again starts charging, the current through inductor starts decreasing, becomes zero, changes direction and again starts increasing and goes beyond the value of 5 A.
Current through the inductor will reach maximum value when charge on the capacitor has become zero.
From conservation of energy between initial state (when i = 5 A, q = 40 μC) and final state (when i = I_{max}, q = 0) we get
A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shape assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed v_{o} and the radius of the cross section of the liquid stream is r_{o}.
Q.
An equation for the speed v of the liquid as a function of the distance y it has fallen, is
Applying Bernoulli’s equation between the initial crosssection and the crosssection at a distance ‘y’ below the pipe we get,
Since both crosssections are in atmosphere, the pressure there is atmospheric pressure,
A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shape assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed v_{o} and the radius of the cross section of the liquid stream is r_{o}.
Q.
The expression for the radius of the crosssection of the liquid stream at a distance ‘y’ below the pipe is
Applying equation of continuity between the initial crosssection and the cross section at a distance ‘y’ below the pipe, we get
Using the expression for ‘v’ from above question, we get
The correct order of acidic strength of the compounds, 1 – 3 is
The correct order of bond angle of the following compound is
The rate law for one of the mechanism of the pyrolysis of CH_{3}CHO at 520^{o}C and 0.2 bar is
The overall activation energy Ea in terms of rate law is:
From Arrhenius equation, K = Ae^{Ea/RT }
From rate equation given in question
Hence (C) is correct answer.
The stopping potential (V_{0}) of photosensitive surface varies with the frequency of incident radiation as shown in the following figure.
The work function of photosensitive surface.
From photoelectric effect we have
From equation (ii) and figure given in question, we have v_{0} = 4×10^{15}Hz
The work function of photosensitive surface is given by the relation:
5.6 litre of an unknown gas at NTP requires 12.5 calorie to raise its temperature by 10^{o}C at constant volume.
Moles of gas
The value of (PV_{m})_{P→0} of a real gas is independent of the nature of gas because
At very low pressure the molar volume of gas is very high therefore; V_{m} >>>b , and molecular attraction are also negligible. Thus gas behaves as an ideal gas.
Hence (A) is correct answer.
Which of the following oxide cannot be reduced by carbon reduction?
MnO_{2} and Cr_{2}O_{3} are reduced by aluminum.
Which of the following can show both optical and geometrical isomeris?
[Co(NH_{3})_{3}(SO_{4})(NO_{2})Cl] forms four geometrical isomers (two cis and two trans) one of the cisisomer is optically active.
Cis isomer is optically active due absence of plane of symmetry
Which of the following is (are) correct statement (s)?
A is correct because nitrogen in +3 oxidation state.
HOF is unstable but acts as very strong oxidizing agent.
moles of CH_{3}COONa = 150 x 0.1 = 15
moles of HCl = 50 x 0.1 = 5
moles of CH3COONa unreacted = 10
moles of weak acid CH_{3}COOH = 5
hence solution is an acidic mixture.
(D) is correct because pH of the salts = (pKa_{1}+pKa_{2})/2
The presence of electron donating group decreases the reactivity of benzaldehyde derivatives towards Cannizzaro’s reaction.
Bond ord er of , SO_{2} and C_{6}H_{6} is 1.5, and bond order of O_{2} is 2.
A reversible process is performed in such a way that the system passes from state 1 to state 2 by pathI and then from state 2 to state one by pathII as shown in figure 5. For the above process
That the quantity is not depends on the path of the reaction, but depends on initial and final state of the system.
Q.
The quantity in the above process is called
A reversible process is performed in such a way that the system passes from state 1 to state 2 by pathI and then from state 2 to state one by pathII as shown in figure 5. For the above process
That the quantity is not depends on the path of the reaction, but depends on initial and final state of the system.
Q.
In thermodynamics, a process is called reversible when:
A reversible process is performed in such a way that the system passes from state 1 to state 2 by pathI and then from state 2 to state one by pathII as shown in figure 5. For the above process
That the quantity is not depends on the path of the reaction, but depends on initial and final state of the system.
Q.
Observe the given curve for a substance and identify the correct statement.
Entropy follows the order of gas > liquid > solid
There are some deposits of nitrates and phosphates in the earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions.
Q.
Select the correct statement among the following:
Almost all metal nitrates are soluble in water
There are some deposits of nitrates and phosphates in the earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions.
Nitrates are mainly produced for use as fertilizers in agriculture because of their
The degree of unsaturation in the product of following reaction is
Which chlorine in the following compound is most reactive towards nucleophilic (weak base) substitution reaction?
Chlorine (3) gives IPSO substitution reaction more easily because of –R effect of nitro group.
Range of the function,
, (where sgn (⋅) denotes the signum function and [⋅] greatest integer function, {.} is a fractional function)
Let the circles S_{1} ≡ x^{2} + y^{2} – 4x – 8y + 4 = 0 and S_{2} be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S_{2} is
Centre of circle S_{1} = (2, 4)
Centre of circle S_{2} = (4, 2)
Radius of circle S_{1} = radius of circle S_{2} = 4
∴ equation of circle S_{2}
(x – 4)^{2} + (y – 2)^{2} = 16
⇒ x^{2} + y^{2} – 8x – 4y + 4 = 0 . . . (i)
Equation of circle touching y = x at (1, 1) can be taken as
(x – 1)^{2} + (y – 1)^{2} + λ(x – y) = 0
or, x^{2} + y^{2} + x (λ – 2) + y(– λ – 2) + 2 = 0 . . . (ii)
As this is orthogonal to S_{2}
Let the line x + y = meets the lines y = xtanθ; θ = 0, at A, B, C and D respectively. If AB, BC and CD are in A.P., then α =
Since AB, B C and CD are in A.P. and let their lengths be taken as a – d, a and a + d respectively.
Three parallel chords of a circle have lengths 2, 3, 4 and subtend angles α, β, α + β at the centre respectively (given α + β < π), then cos α is equal to
Using the property that equal chords subtends equal angles at centre of circle, then problem can be converted to the diagram in adjoining figure.
AB = 4, AC = 2, BC = 3
Apply Cosine rule in ΔABC,
If α, β , γ are the roots of the equation x^{3} = x^{2} + 1, then the equation whose roots are α^{2} + β^{3} + γ^{4}, β^{2} + γ^{3} + α^{4}, γ^{2} + α^{3} + β^{4} is
be a point and be a line. If PQ is the distance of plane from point P measured along a line inclined at an angle of 60^{o} with the L and is minimum, then PQ is equal to
Required distance PQ is perpendicular distance of plane from P (as angle between line and plane is 30^{o})
Put x = 0
⇒ f(y^{2}) = f^{2}(0) + y^{2} (1)
Taking y = 0 in (1), we get
f(0) = f^{2}(0)
⇒ f(0) = 0 or f(0) = 1
Now, taking y = x in the given equation
f(0) = f^{2}(x) – 2xf(x) + x^{2} (2)
Case I: If f(0) = 0, then f(x) = x.
Case II: If f(0) = 1, then f(x) = x ± 1
hence f(x) = x + 1 (only)
If polynomials f(x), g(x) and h(x), are such that
Q.
f(x) =
The function G (x) = max .{−1, 3x + 2} agrees with F(x) for x ≤0 , but
If polynomials f(x), g(x) and h(x), are such that
Q.
g(0) =
The function G (x) = max .{−1, 3x + 2} agrees with F(x) for x ≤0 , but
If polynomials f(x), g(x) and h(x), are such that
Q.
h' (0) =
The function G (x) = max .{−1, 3x + 2} agrees with F(x) for x ≤0 , but
P, Q are two points on the rectangular hyperbola (x – 1)(y – 2) = c^{2}, O is the centre of hyperbola, also tangent at P is perpendicular to OQ and meets OQ at N such that (OQ)(ON) = 4
Q.
One of the equation of directrix of the hyperbola is
so equation of hyperbola is (x – 1)(y – 2) = 2
Its foci are (−1, 0), (3, 4) and equation of directrix is x + y – 5 = 0, x + y – 7 = 0 is the equation of latus rectum of ellipse and tangents at end points of latus rectum always intersects on corresponding directrix.
P, Q are two points on the rectangular hyperbola (x – 1)(y – 2) = c^{2}, O is the centre of hyperbola, also tangent at P is perpendicular to OQ and meets OQ at N such that (OQ)(ON)=4
Q.
An ellipse confocal with the hyperbola and with eccentricity equal to is intersected by the line x + y – 7 = 0 at A and B, then intersection point of tangents at A and B will lie on
so equation of hyperbola is (x – 1)(y – 2) = 2
Its foci are (−1, 0), (3, 4) and equation of directrix is x + y – 5 = 0, x + y – 7 = 0 is the equation of latus rectum of ellipse and tangents at end points of latus rectum always intersects on corresponding directrix.
‘A’ tosses a coin. If it shows a tail, ‘A’ is asked to roll a fair die and A’s score is the number that die shows. If the coin shows a head, ‘A’ is asked to toss five more coins and A’s score is that total number of heads shown including the first coin. If A tells you only that his score is 3, and if ‘P’ is the probability that ‘A’ rolled a die, then [1/P] is equal to (where [ ] is the greatest integer function)
Let us define the events as follows:
E_{1} : A rolls the die
E_{2} : A tosses five more coins
E : A’s score is 3
The number of solution of the curve sin x = cos y and the circle x^{2} + y^{2} = 1 is/are
Equation of circle in parametric form
x = cos θ, y = sin θ
So, we have to solve the equation sin (cos θ) = cos (sin θ) (1)
Which is not satisfied by any n = 0, ±1, ± 2, ......
∴ sin x = cos y does not intersect the circle x^{2} + y^{2} = 1.
2 videos327 docs203 tests

2 videos327 docs203 tests
