Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B?

  • 4.9 ms-2 in horizontal direction

  • 9.8 ms-2 in vertical direction

  • zero

  • 4.9 ms-2 in vertical direction


D.

4.9 ms-2 in vertical direction

For the motion of block along inclined plane
mg sin θ =ma
a = g sin θ
where a is along the inclined plane.
The vertical component of acceleration is g sin2θ
Therefore, the relative vertical acceleration of  A with respect to B is 
g (sin260 - sin230) = g/2 = 4.9 ms-2 (in vertical direction)

453 Views

2.

Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

  • 100N

  • 80 N

  • 120 N

  • 150 N


C.

120 N

In the vertical direction, weight are balanced by frictional forces.
As the blocks are in equilibrium balance forces are in horizontal and vertical direction.
For the system of blocks (A+B)
F = N
For block A, fA = 20 N and for block B.
fB = fA +100 = 120 N

731 Views

3.

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

  • 44%

  • 50%

  • 56%

  • 62%


C.

56%

354 Views

4.

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

  • g

  • 2/3g

  • g/3

  • 3/2g


B.

2/3g


mg - T = ma
TR space equals space fraction numerator mR squared straight alpha over denominator 2 end fraction
straight T space equals space mRα over 2 space equals space ma over 2
mg space minus ma over 2 space equals space ma
fraction numerator 3 space ma over denominator 2 end fraction space equals space mg
straight a space equals space fraction numerator 2 straight g over denominator 3 end fraction
514 Views

5.

A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

  • 2g/3

  • g/2

  • 5g/6

  • g


B.

g/2

For the mass m,
mg-T = ma


As we know, a = Rα    ... (i)
So, mg-T = mRα
Torque about centre of pully
T x R = mR2α ...... (ii)
From Eqs. (i) and (ii), we get 
a = g/2
Hence, the acceleration of the mass of a body fall is g/2.

455 Views

6.

STATEMENT – 1
Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
STATEMENT – 2
Principle of conservation of momentum holds true for all kinds of collisions.

  • The statement I is True, Statement II is False.

  • The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

  • The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.

  • Statement I is False, Statement II is


B.

The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

264 Views

7.

A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ0) and t is


A.

According to Newton's law of cooling.

dθ over dt space proportional to left parenthesis straight theta minus straight theta subscript straight o right parenthesis
rightwards double arrow fraction numerator begin display style dθ end style over denominator begin display style dt end style end fraction space equals space minus straight k left parenthesis straight theta minus straight theta subscript straight o right parenthesis
integral fraction numerator begin display style dθ end style over denominator begin display style straight theta minus straight theta subscript straight o end style end fraction space equals space integral negative kdt space
rightwards double arrow space In space left parenthesis straight theta minus straight theta subscript straight o right parenthesis space equals space minus kt plus straight c

Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.

473 Views

8.

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10–2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is

  • 0.0125 Nm-1

  • 0.1 Nm-1

  • 0.05 Nm-1

  • 0.025 Nm-1


D.

0.025 Nm-1



The force of surface tension acting on the slider balances the force due to the weight.

⇒F = 2Tl = w
⇒2T(0.3) = 1.5 x 10–2
⇒T = 2.5 x 10–2 N/m

369 Views

9.

A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above ground at which the block can be placed without slipping is 

  • 1 over 6 straight m
  • 2 over 3 straight m
  • 1 third straight m
  • 1 half straight m

A.

1 over 6 straight m

A block of mass m is placed on a surface with vertical cross section, then


tan space straight theta space equals space dy over dx space fraction numerator straight d open parentheses begin display style straight x cubed over 6 end style close parentheses over denominator dx end fraction space equals space straight x squared over 2
At limiting equilibrium, we get
μ = tan θ
0.5 = x2/2
⇒ x2 =1
⇒ x = ±1
Now, putting the value of x in y = x3/6, we get
When x =1

straight y space equals space fraction numerator left parenthesis 1 right parenthesis cubed over denominator 6 end fraction space equals 1 over 6
When x =-1

straight y space equals space fraction numerator left parenthesis negative 1 right parenthesis cubed over denominator 6 end fraction space equals fraction numerator negative 1 over denominator 6 end fraction
So, the maximum height above the ground at which the block can be placed without slipping is 1/6m.

234 Views

10.

A pulley of radius 2m is rotated about its axis by a force F = (20t - 5t2) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, the number of rotations made by the pulley before its direction of motion if reversed is

  • more than 3 but less than 6

  • more than 6 but less than 9

  • more than 9

  • less than


A.

more than 3 but less than 6

To reverse the direction
integral τdθ space equals 0
straight tau space equals space left parenthesis 20 space straight t minus 5 straight t squared right parenthesis 2 space equals space 40 straight t minus 10 straight t squared
straight alpha space equals space straight tau over straight I space equals space fraction numerator 40 straight t minus 10 straight t squared over denominator 10 end fraction space equals space 4 straight t minus straight t squared
straight omega space equals space integral subscript 0 superscript straight t space αdt space equals space 2 straight t squared minus straight t cubed over 3
straight omega space is space zero space at
2 straight t squared minus straight t cubed over 3 space equals space 0
straight t cubed space equals space 6 straight t squared
straight t space equals space 6 space sec
straight theta space equals space integral ωdt
space equals space integral subscript 0 superscript 6 left parenthesis 2 straight t squared minus straight t cubed over 3 right parenthesis dt
open square brackets fraction numerator 2 straight t cubed over denominator 3 end fraction minus straight t to the power of 4 over 12 close square brackets subscript 0 superscript 6 space equals space 216 space open square brackets 2 over 3 minus 1 half close square brackets space equals space 36 space rad.
No space of space revolution space fraction numerator 36 over denominator 2 straight pi end fraction space Less space than space 6

869 Views