Free Objective Test 02 Practice Test - 11th and 12th

Question 1

A weightless ladder, 20 ft long rests against a frictionless wall at an angle of 60⁰ with the horizontal. A 150 pound main is 4 ft from the top of the ladder. A horizontal force is needed to prevent it from slipping choose the correct magnitude from the following

175 lb

100 lb

70 lb

150 lb

SOLUTION

Solution : C

Since the system is in equilibrium therefore $\sum F_{x} = 0 a n d \sum F_{y} = 0 ∴ F = R_{2} a n d W = R_{1}$ Now by taking the moment of forces about point B.

$F . (B C) + W . (E C) = R_{1}$ (AC) [from the figure EC = 4 cos 60

$F . (20 s i n 60) + W (4 c o s 60) = R_{1} (20 c o s 60)$

$10 \sqrt{3} F + 2 W = 10 R_{1} [A s R_{1} = W] ∴ F = \frac{8 W}{10 \sqrt{3}} = \frac{8 \times 150}{10 \sqrt{3}} = 701 b$

Question 2

A mass M is suspended by a rope from a rigid support at P as shown in the figure. Another rope is tied at the end Q, and it is pulled horizontally with a force F. If the rope PQ makes angle $θ$ with the vertical then the tension in the string PQ is

$F s i n θ$

$\frac{F}{s i n θ}$

$C c o s θ$

$\frac{F}{c o s θ}$

SOLUTION

Solution : B

From the figure

For horizontal equilibrium

$T s i n θ = F$

$∴ T = \frac{F}{s i n θ}$

Question 3

A spring balance A shows a reading of 2 kg, when an aluminium block is suspended from it. Another balance B shows a reading of 5 kg, when a beaker full of liquid is placed in its pan. The two balances are arranged such that the Al – block is completely immersed inside the liquid as shown in the figure. Then

The reading of the balance A will be more than 2 kg

The reading of the balance B will be less than 5 kg

The reading of the balance A will be less than 2 kg. and that of B will be more than 5 kg

The reading of balance A will be 2 kg. and that of B will be 5 kg.

SOLUTION

Solution : C

Due to buoyant force on the aluminium block the reading of spring balance A will be less than 2 kg but it increase the reading of balance B.

Question 4

A man of mass m stands on a create of mass M. He pulls on a light rope passing over a smooth light pulley. The other end of the rope is attached to the crate. For the system to be in equilibrium, the force exerted by the men on the rope will be

$(M + m) g$

$\frac{1}{2} (M + m) g$

$M g$

$m g$

SOLUTION

Solution : B

From the free body diagram of man and crate system: For vertical equilibrium

2T = (M + m)g

$∴ T = \frac{(M + m)}{2} g$

Question 5

A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01s on the ball. The average force exerted by the bat on the ball is

480 N

600 N

500 N

400 N

SOLUTION

Solution : A

$v_{1} = - 12$ m\s and $v_{2} = + 20$ m/s [because direction is reversed]

$m = 150 g m = 0.15 k g, t = 0.01 s e c$

Force exerted by the bat on the ball $F = \frac{m [v_{2} - v_{1}]}{t} = \frac{0.15 [20 - (- 12)}{0.01}] = 480 N e w t o n$

Question 6

The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration ‘a’ is 3:2. The value of ‘a’ is (g – Acceleration due to gravity on the earth)

$\frac{3}{2} g$

$\frac{g}{3}$

$\frac{2}{3} g$

$g$

SOLUTION

Solution : B

$\frac{w e i g h t o f a m a n i n s t a t i o n a r y l i f t}{w e i g h t o f a m a n i n d o w n w a r d m o v i n g l i f t} = \frac{m g}{m (g - a)} = \frac{3}{2} \frac{g}{g - a} = \frac{3}{2} \Rightarrow 2 g = 3 g - 3 a o r a = \frac{g}{3}$

Question 7

A 60 kg man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from 60 kg to 50 kg for a while and then comes back to the original mark. What should we conclude

The lift was in constant motion upwards

The lift was in constant motion downwards

The lift while in constant motion upwards, is stopped suddenly

The lift while in constant motion downwards, is suddenly stopped

SOLUTION

Solution : C

For retarding motion of a lift R = m(g + a) for downward motion

R = m(g – a) for upward motion

Since the weight of the body decrease for a while and then comes back to original value it means the lift was moving upward and stops suddenly.

Note: Generally we use R = m(g + a) for downward motion

R = m(g – a) for upward motion

Here a = acceleration, but for the given problem a = retardation

Question 8

A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight placed on a spring balance, It records a weight of 25 N. The bird (mass = 0.5kg) flies upward in the cage with an acceleration of $2 m / s^{2}$ . The spring balance will now record a weight of

24 N

25 N

26 N

27 N

SOLUTION

Solution : B

Since the cage is closed and we can treat bird cage and air as a closed (Isolated ) system. In this condition the force applied by the bird on the cage is an internal force due to this reading of spring balance will not change.

Question 9

A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be $W_{1}$ .If the bird flies about inside the cage, the reading of the spring balance is $W_{2}$ . Which of the following is true

$W_{1} = W_{2}$

$W_{1} > W_{2}$

$W_{1} < W_{2}$

D. Nothing definite can be predicted

SOLUTION

Solution : A

Again, the bird is not flying because of the spring force, or normal forse from the cage, right! So, $W_{1} = W_{2}$

Question 10

Two blocks of mass 4 kg and 6 kg are placed in contact with each other on a frictionless horizontal surface. If we apply a push of 5 N on the heavier mass, the force on the lighter mass will be

5 N

4 N

2 N

None of the above

SOLUTION

Solution : C

$L e t m_{1} = 6 k g, m_{2} = 4 k g a n d F = 5 N (g i v e n)$ Force on the lighter mass $= \frac{m_{2} \times F}{m_{1} + m_{2}} = \frac{4 \times 5}{6 + 4} = 2 N$

Question 11

The acceleration of block B in the figure will be

\frac{m_{2} g}{(4 m_{1} + m_{2})}

\frac{2 m_{2} g}{(4 m_{1} + m_{2})}

\frac{2 m_{1} g}{(m_{1} + 4 m_{2})}

\frac{2 m_{1} g}{(m_{1} + m_{2})}

SOLUTION

Solution : A

When the block $m_{2}$ moves downward with acceleration a, the acceleration of mass $m_{1}$ will be 2a because it covers double distance in the same time in comparison to $m_{2}$ .
Let T is the tension in the string.
By drawing the free body diagram of A and B
$T = m_{1} 2 a - - - - - - - - - - (i)$
$M_{2} g - 2 T = m_{2} a - - - - - - - - - - - (i i)$
By solving (i) and (ii)
By solving (i) and (ii)
$a = \frac{m_{2} g}{(4 m_{1} + m_{2})}$

Question 12

In the above problem, if the speed of $m_{2}$ is 50 cm/s the distance covered by $m_{1}$ in 0.4 s will be

A. 40 cm

B. 20 cm

C. 10 cm

D. 80 cm

SOLUTION

Solution : A

Distance covered by $m_{2} = 50 c m / s \times 0.4 s = 20 c m$ Since the $m_{1}$ mass cover double distance therefore $S = 2 \times 20 = 40 c m$

Question 13

In the above problem, if we consider the two masses to be a system, the tension is an external force.

A. True

B. False

SOLUTION

Solution : B

Tension is the internal force in this case. Gravitational force is the only external force.

Question 14

A body weighs 8 gm, when placed in one pan and 18gm, when placed in the other pan of a false balance. The beam is horizontal (when both the pans are empty), the true weight of the body is

A. 13 gm

B. 12 gm

C. 15.5 gm

D. 15 gm

SOLUTION

Solution : B

For given condition true weight $= \sqrt{W_{1} W_{2}} = \sqrt{8 \times 18} = 12 g m$ .

Question 15

At what minimum acceleration should a monkey slide down a rope whose breaking strength is two-thirds of the monkey's weight?

A. g

\frac{g}{3}

\frac{2 g}{3}

D. 2g

SOLUTION

Solution : B

When a monkey is sliding down a rope, the net force acting on it is given by
ma = mg - T, where T is the breaking strength of the rope
$\Rightarrow$ ma = mg - $\frac{2}{3}$ mg
$\Rightarrow$ a = $\frac{g}{3}$

Free Objective Test 02 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

Related ExamsObjective Test 02Objective Test 02Objective Test 02

Related Exams
Objective Test 02
Objective Test 02
Objective Test 02