Free Objective Test 01 Practice Test - 11th and 12th

Question 1

A force $F = - K (y i + a j)$ (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is

$- 2 K a^{2}$

$2 K a^{2}$

$- K a^{2}$

$K a^{2}$

SOLUTION

Solution : C

While moving from (0,0) to (a,0)

Along positive x-axis, y = 0 $∴ \to F = - k x^j$

i.e. force is in negative y-direction while displacement is in positive x-direction.

$∴ W_{1} = 0$

Because force is perpendicular to displacement

Then particle moves from (a,0) to (a,a) along a line parallel to y-axis (x = +a) during this $\to F = - k (y^i + a^j)$

The first component of force, $- k y^i$ will not contribute any work because this components is along negative x-direction ( $-^i)$ while displacement is in positive y-direction (a,0) to (a,a).

The second component of force i.e. $- k a^j$

$∴ W_{2} = (- k a^j) (a^j) = (- k a) (a) = - k a^{2}$

So net work done on the particle $W = W_{1} + W_{2}$

$= 0 + (- k a^{2}) = - k a^{2}$

Question 2

Which of the following is a scalar quantity

Displacement

Electric field

Acceleration

Work

SOLUTION

Solution : D

Work is a dot product of force and displacement and hence a scalar.

Question 3

If the unit of force and length each be increased by four times, then the unit of energy is increased by

16 times

8 times

2 times

4 times

SOLUTION

Solution : A

Work = Force $\times$ Displacement

If unit of force and length be increased by four times then the unit of energy will increase by 16 times.

Question 4

A man pushes a wall and fails to displace it. He does

Negative work

Positive but not maximum work

No work at all

Maximum work

SOLUTION

Solution : C

No displacement is there. Hence work is zero.

Question 5

When work is done on a body by an external force, its

Only kinetic energy increases

Only potential energy increases

Both kinetic and potential energies may increase

Sum of kinetic and potential energies remains constant

SOLUTION

Solution : C

Work done will convert into energies - both Kinetic and Potential

Question 6

The work done in pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is

4.36 kJ

5.17 kJ

8.91 kJ

9.82 kJ

SOLUTION

Solution : B

$W = m g s i n θ \times s$

$= 2 \times 10^{3} \times s i n 15^{\circ} \times 10$

$= 5.17 k J$

Question 7

A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

5.28 J

450 mJ

490 mJ

530 mJ

SOLUTION

Solution : A

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

$∴ v_{0} = 3 m / s a n d v_{4} = 19 m / s$

$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$ (According to work energy theorem)

$= \frac{1}{2} \times 0.03 \times (19^{2} - 3^{2}) = 5.28 J$

Question 8

A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25 joule, the angle which the force makes with the direction of motion of the body is

$0^{\circ}$

$30^{\circ}$

$60^{\circ}$

$90^{\circ}$

SOLUTION

Solution : C

$W = F s c o s θ$
$\Rightarrow c o s θ = \frac{W}{F s} = \frac{25}{50} = \frac{1}{2}$
$\Rightarrow θ = c o s^{- 1} (\frac{1}{2})$
$\Rightarrow θ = 60^{\circ}$

Question 9

A body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is $\frac{m v^{2}}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle

$\frac{m v^{2}}{π r^{2}}$

Zero

$\frac{m v^{2}}{r^{2}}$

$\frac{π r^{2}}{m v^{2}}$

SOLUTION

Solution : B

Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular.

$W = \to F . \to s = F s c o s θ = F s c o s 90^{\circ} = 0$

Question 10

Given $| \to A + \to B | = | \to A - \to B |$ Find the angle between $\to A$ and $\to B$

$0^{0}$

$45^{0}$

$90^{0}$

$60^{0}$

SOLUTION

Solution : C

$A^{2} + B^{2} + 2 A B c o s θ = A^{2} + B^{2} - 2 A c o s θ \Rightarrow 4 A B c o s θ = 0 \Rightarrow c o s θ = 0 \Rightarrow θ = 90^{\circ}$

Question 11

A moving body of mass m and velocity 3 km/h collides with a body at rest of mass 2m and sticks to it. Now the combined mass starts to move. What will be the combined velocity

3 km/h

2 km/h

1 km/h

4 km/h

SOLUTION

Solution : C

Initial momentum = $m \times 3 + 2 m \times 0 = 3 m$

Final momentum = $3 m \times V$

By the law of conservation of momentum

$3 m = 3 m \times V \Rightarrow V = 1 k m / h$

Question 12

The kinetic energy acquired by a mass m in travelling a certain distance d, starting from rest, under the action of a force F such that the force F is directly proportional to t is

Directly proportional to $t^{2}$

Independent of t

Directly proportional to $t^{4}$

Directly proportional to t

SOLUTION

Solution : C

f = kt

$m \int_{0}^{v}$ dv = $k \int_{0}^{t}$ tdt
mv = k $t^{2}$
v = $\frac{k}{m}$ $t^{2}$
KE = $\frac{1}{2}$ m $v^{2}$ = $\frac{1}{2}$ m $\frac{k^{2}}{m^{2}}$ $t^{4}$ = $($ $\frac{k^{2}}{2 m}$ $)$ $t^{4}$
KE $α$ $t^{4}$

Question 13

The blocks A and B shown in fig. Have masses $M_{A} = 5 k g$ and $M_{B} = 4 k g$ . The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is

$\frac{\sqrt{3}}{2} \sqrt{g}$

$\frac{\sqrt{3}}{4} \sqrt{g}$

$\frac{\sqrt{g}}{2 \sqrt{3}}$

$\frac{\sqrt{g}}{2}$

SOLUTION

Solution : C

If A moves down the incline by 1 m, B shal move up by $\frac{1}{2} m$ .

If the speed of B is v, then the speed of A will be 2v.

From conservation of energy,

Gain in KE = loss in PE

$\frac{1}{2} m_{A} (2 v)^{2} + \frac{1}{2} m_{B} v^{2} = m_{A} g \times \frac{3}{5} - m_{B} g \times \frac{1}{2}$

Solving, we get

$v = \frac{\sqrt{g}}{2 \sqrt{3}}$

Question 14

The velocity - time graph of a particle moving in a straight line is shown in figure. The mass of the particle is 2kg.Work done by all the forces acting on the particle in time interval between t = 0 to t = 10 s is

300 J

−300 J

400 J

−400 J

SOLUTION

Solution : A

From work - energy theorem,

W = $△ K E = K_{f} - K_{i} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$

= $\frac{1}{2} \times 2 [(- 20)^{2} - (10)^{2}] = 300 J$

Question 15

The kinetic energy acquired by a mass m in travelling a certain distance d starting from rest under the action of a constant force is directly proportional to

$\sqrt{m}$

Independent of m

$\frac{1}{\sqrt{m}}$

SOLUTION

Solution : B

Kinetic energy acquired by the body

= Force applied on it $\times$ Distance covered by the body

$K . E = F \times d$

If F and d both are same then K.E. acquired by the body will be same

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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