Work, Power & Energy
769 Questions
Start Allen Test
JEE (Main) June 2022
Q51
Errorless
Critical Thinking
MCQ
What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass (Assume the collision to be head-on elastic collision)
A.
$50.0\%$
B.
$66.6\%$
C.
$55.5\%$
D.
$33.3\%$
JEE (Main) June 2022
Q52
Errorless
Critical Thinking
MCQ
A particle of mass 500 gm is moving in a straight line with velocity $v = bx^{5/2}$ . The work done by the net force during its displacement from x = 0 to x = 4m is (Take $b = 0.25 m^{-3/2} s^{-1}$ )
A.
2 J
B.
4 J
C.
8 J
D.
16 J
JEE (Main) June 2022
Q53
Errorless
Critical Thinking
MCQ
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as $a = k^{2}rt^{2}$ . Where k is a constant. The power delivered to the particle by the force acting on it is given as
A.
Zero
B.
$mk^{2}r^{2}t^{2}$
C.
$mk^{2}r^{2}t$
D.
$mk^{2}rt$
JEE (Main) July 2022
Q54
Errorless
Critical Thinking
MCQ
A bullet of mass $200g$ having initial kinetic energy $90J$ is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to $40J$ within 1s, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is
A.
45 m
B.
90 m
C.
125 m
D.
25 m
JEE (Main) July 2022
Q55
Errorless
Critical Thinking
MCQ
A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed $10 \, ms^{-1}$ gets embedded in it, then loss of kinetic energy will be
A.
4.9 J
B.
9.8 J
C.
14.7 J
D.
19.6 J
JEE (Main) July 2022
Q56
Errorless
Critical Thinking
MCQ
As per the given figure, two blocks each of mass 250g are connected to a spring of spring constant $2Nm^{-1}$ . If both are given velocity v in opposite directions, then maximum elongation of the spring is
A.
$\frac{v}{2\sqrt{2}}$
B.
$\frac{v}{2}$
C.
$\frac{v}{4}$
D.
$\frac{v}{\sqrt{2}}$
JEE (Main) July 2022
Q57
Errorless
Critical Thinking
MCQ
A body of mass 8 kg and another of mass 2 kg are moving with equal kinetic energy. The ratio of their respective momenta will be
A.
1:1
B.
2:1
C.
1:4
D.
4:1
JEE (Main) July 2022
Q58
Errorless
Critical Thinking
MCQ
If momentum of a body is increased by 20%, then its kinetic energy increases by
A.
36%
B.
40%
C.
44%
D.
48%
JEE (Main) July 2022
Q59
Errorless
Critical Thinking
MCQ
Sand is being dropped from a stationary dropper at a rate of $0.5 \, kgs^{-1}$ on a conveyor belt moving with a velocity of $5 \, ms^{-1}$ . The power needed to keep belt moving with the same velocity will be
A.
1.25 W
B.
2.5 W
C.
6.25 W
D.
12.5 W
JEE (Main) June 2022
Q60
Errorless
Graphical Questions
MCQ
Arrange the four graphs in descending order of total work done; where $W_{1}, W_{2}, W_{3}$ and $W_{4}$ are the work done corresponding to figure 1, 2, 3 and 4 respectively
Figure-1
$F\uparrow$
Figure-2
$F\uparrow$
Figure-3
$F\uparrow$
Figure-4
$F\uparrow$
Figure-2
$F\uparrow$
Figure-3
$F\uparrow$
Figure-4
$F\uparrow$
A.
$W_{3} > W_{2} > W_{1} > W_{4}$
B.
$W_{3} > W_{2} > W_{4} > W_{1}$
C.
$W_{2} > W_{3} > W_{4} > W_{1}$
D.
$W_{2} > W_{3} > W_{1} > W_{4}$
JEE (Main) July 2022
Q61
Errorless
Graphical Questions
MCQ
A bullet is shot vertically downwards with an initial velocity of 100 m/s from a certain height. Within 10 s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time t = 20 s will be (Take $g = 10 \, m/s^{2}$ )
A.
a
B.
b
C.
c
D.
d
JEE (Main) 2021
Q62
Allen
JEE Main PYQ
Numerical
A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5:1. The velocity of the bob at the highest position is ____ m/s. (Take g = 10 m/s $^{2}$ )
Correct Answer: 5
Explanation:
Ans. (5)
Let the speed of bob at lowest position be $v_{1}$ and at the highest position be $v_{2}$ . Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,
$\begin{array}{r l r} & {\frac {1}{2} m v _ {1} ^ {2} = \frac {1}{2} m v _ {2} ^ {2} + \mathrm{mg} 2 l} \\ {\Rightarrow} & {v _ {1} ^ {2} = v _ {2} ^ {2} + 4 g l} & {\dots (1)} \\ {\mathrm{Now}} & {T _ {m a x} - m g = \frac {m v _ {1} ^ {2}}{l}} \\ {\Rightarrow} & {T _ {m a x} = m g + \frac {m v _ {1} ^ {2}}{l}} \\ {\mathrm{And}} & {T _ {m i n} + m g = \frac {m v _ {2} ^ {2}}{l}} \\ {\Rightarrow} & {T _ {m i n} = \frac {m v _ {2} ^ {2}}{l} - m g} \\ & {\frac {T _ {m a x}}{T _ {m i n}} = \frac {5}{1} \Rightarrow \frac {m g + \frac {m v _ {1} ^ {2}}{l}}{\frac {m v _ {2} ^ {2}}{l} - m g} = \frac {5}{1}} \\ {\Rightarrow} & {m g + \frac {m v _ {1} ^ {2}}{l} = \left[ \frac {m v _ {2} ^ {2}}{l} - m g \right] 5} \\ {\Rightarrow} & {m g + \frac {m}{l} [ v _ {2} ^ {2} + 4 g l ] = \frac {5 m v _ {2} ^ {2}}{l} - 5 m g} \\ {\Rightarrow} & {m g + \frac {m _ {2} ^ {2}}{l} + 4 m g = \frac {5 m v _ {2} ^ {2}}{l} - 5 m g} \\ {\Rightarrow} & {1 0 m g = \frac {4 m v _ {2} ^ {2}}{l}} \\ & {V _ {2} ^ {2} = \frac {1 0 \times 1 0 \times 1}{4}} \\ {\Rightarrow} & {v _ {2} ^ {2} = 2 5} \\ {\Rightarrow} & {v _ {2} = 5 m / s} \end{array}$
Thus, velocity of bob at highest position is 5 m/s.
Let the speed of bob at lowest position be $v_{1}$ and at the highest position be $v_{2}$ . Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,
$\begin{array}{r l r} & {\frac {1}{2} m v _ {1} ^ {2} = \frac {1}{2} m v _ {2} ^ {2} + \mathrm{mg} 2 l} \\ {\Rightarrow} & {v _ {1} ^ {2} = v _ {2} ^ {2} + 4 g l} & {\dots (1)} \\ {\mathrm{Now}} & {T _ {m a x} - m g = \frac {m v _ {1} ^ {2}}{l}} \\ {\Rightarrow} & {T _ {m a x} = m g + \frac {m v _ {1} ^ {2}}{l}} \\ {\mathrm{And}} & {T _ {m i n} + m g = \frac {m v _ {2} ^ {2}}{l}} \\ {\Rightarrow} & {T _ {m i n} = \frac {m v _ {2} ^ {2}}{l} - m g} \\ & {\frac {T _ {m a x}}{T _ {m i n}} = \frac {5}{1} \Rightarrow \frac {m g + \frac {m v _ {1} ^ {2}}{l}}{\frac {m v _ {2} ^ {2}}{l} - m g} = \frac {5}{1}} \\ {\Rightarrow} & {m g + \frac {m v _ {1} ^ {2}}{l} = \left[ \frac {m v _ {2} ^ {2}}{l} - m g \right] 5} \\ {\Rightarrow} & {m g + \frac {m}{l} [ v _ {2} ^ {2} + 4 g l ] = \frac {5 m v _ {2} ^ {2}}{l} - 5 m g} \\ {\Rightarrow} & {m g + \frac {m _ {2} ^ {2}}{l} + 4 m g = \frac {5 m v _ {2} ^ {2}}{l} - 5 m g} \\ {\Rightarrow} & {1 0 m g = \frac {4 m v _ {2} ^ {2}}{l}} \\ & {V _ {2} ^ {2} = \frac {1 0 \times 1 0 \times 1}{4}} \\ {\Rightarrow} & {v _ {2} ^ {2} = 2 5} \\ {\Rightarrow} & {v _ {2} = 5 m / s} \end{array}$
Thus, velocity of bob at highest position is 5 m/s.
JEE (Main) 2021
Q63
Allen
JEE Main PYQ
Numerical
A ball of mass 4 kg, moving with a velocity of $10 \, ms^{-1}$ , collides with a spring of length 8 m and force constant $100 \, Nm^{-1}$ . The length of the compressed spring is x m. The value of x, to the nearest integer, is ____.
Correct Answer: 6
Explanation:
Ans. (6)
Let's say the compression in the spring be y.
So, by work energy theorem we have
$\Rightarrow \frac {1}{2} m v ^ {2} = \frac {1}{2} k y ^ {2}$
$\Rightarrow y = \sqrt {\frac {m}{k}} \cdot v$
$\Rightarrow y = \sqrt {\frac {4}{1 0 0}} \times 1 0$
Let's say the compression in the spring be y.
So, by work energy theorem we have
$\Rightarrow \frac {1}{2} m v ^ {2} = \frac {1}{2} k y ^ {2}$
$\Rightarrow y = \sqrt {\frac {m}{k}} \cdot v$
$\Rightarrow y = \sqrt {\frac {4}{1 0 0}} \times 1 0$
JEE (Main) 2021
Q64
Allen
JEE Main PYQ
Numerical
In a spring gun having spring constant 100 N/m a small ball 'B' of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground. The value of d is ____ m. (g = 10 m/s²)
Correct Answer: 1
Explanation:
Ans. (1)
$\frac {1}{2} k x ^ {2} = \frac {1}{2} m v ^ {2}$
$k x ^ {2} = m v ^ {2}$
$v = x \sqrt {\frac {k}{m}} = 0.0 5 \sqrt {\frac {1 0 0}{0.1}} = 0.0 5 \times 1 0 \sqrt {1 0}$
$v = 0.5 \sqrt {1 0}$
From $h = \frac{1}{2} gt^2$
$t = \sqrt {\frac {2 h}{g}} = \sqrt {\frac {2 \times 2}{1 0}} = \frac {2}{\sqrt {1 0}}$
$\therefore d = v t = 0.5 \sqrt {1 0} \times \frac {2}{\sqrt {1 0}} = 1 m$
$\frac {1}{2} k x ^ {2} = \frac {1}{2} m v ^ {2}$
$k x ^ {2} = m v ^ {2}$
$v = x \sqrt {\frac {k}{m}} = 0.0 5 \sqrt {\frac {1 0 0}{0.1}} = 0.0 5 \times 1 0 \sqrt {1 0}$
$v = 0.5 \sqrt {1 0}$
From $h = \frac{1}{2} gt^2$
$t = \sqrt {\frac {2 h}{g}} = \sqrt {\frac {2 \times 2}{1 0}} = \frac {2}{\sqrt {1 0}}$
$\therefore d = v t = 0.5 \sqrt {1 0} \times \frac {2}{\sqrt {1 0}} = 1 m$
JEE (Main) 2021
Q65
Allen
JEE Main PYQ
MCQ
If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :
A.
100%
B.
200%
C.
300%
D.
400%
JEE (Main) 2021
Q66
Allen
JEE Main PYQ
MCQ
A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase.
(Take $g = 9.8 \, ms^{-2}$ )
A.
-62720.0 J
B.
-627.2 J
C.
+627.2 J
D.
784.0 J
JEE (Main) 2021
Q67
Allen
JEE Main PYQ
MCQ
A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of $20 \, ms^{-1}$ . The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?
A.
$19.0 \, ms^{-1}$
B.
$4.47 \, ms^{-1}$
C.
$14.41 \, ms^{-1}$
D.
$1.00 \, ms^{-1}$
JEE (Main) 2021
Q68
Allen
JEE Main PYQ
Numerical
A small block slides down from the top of hemisphere of radius $R = 3 \, m$ as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is __m.
(Assume there is no friction between the block and the hemisphere)
Correct Answer: 2
Explanation:
Ans. (2)
$mg \cos \theta = \frac{mv^{2}}{R}$
...(1)
$\cos \theta = \frac{h}{R}$
Energy conservation
$m g \{R - h \} = \frac {1}{2} m v ^ {2}$
...(2)
From (1) and (2)
$m g \Bigl \{\frac {h}{R} \Bigr \} = \frac {2 m g \{R - h \}}{R}$
$h = \frac {2 R}{3} = 2 m$
$mg \cos \theta = \frac{mv^{2}}{R}$
...(1)
$\cos \theta = \frac{h}{R}$
Energy conservation
$m g \{R - h \} = \frac {1}{2} m v ^ {2}$
...(2)
From (1) and (2)
$m g \Bigl \{\frac {h}{R} \Bigr \} = \frac {2 m g \{R - h \}}{R}$
$h = \frac {2 R}{3} = 2 m$
JEE (Main) 2021
Q69
Allen
JEE Main PYQ
Numerical
A uniform chain of length 3 meter and mass 3 kg overhangs a smooth table with 2 meter lying on the table. If k is the kinetic energy of the chain in joule as it completely slips off the table, then the value of k is ____ (Take $g = 10 \, m/s^{2}$ )
Correct Answer: 40
Explanation:
Ans. (40)
From energy conservation
$\begin{array}{l} K _ {i} + U _ {i} = k _ {f} + U _ {f} \\ 0 + \left(- 1 \times 1 0 \times \frac {1}{2}\right) = k _ {f} + \left(- 3 \times 1 0 \times \frac {3}{2}\right) \\ - 5 = k _ {f} - 4 5 \\ k _ {f} = 4 0 J \end{array}$
From energy conservation
$\begin{array}{l} K _ {i} + U _ {i} = k _ {f} + U _ {f} \\ 0 + \left(- 1 \times 1 0 \times \frac {1}{2}\right) = k _ {f} + \left(- 3 \times 1 0 \times \frac {3}{2}\right) \\ - 5 = k _ {f} - 4 5 \\ k _ {f} = 4 0 J \end{array}$
JEE (Main) 2021
Q70
Allen
JEE Main PYQ
Numerical
Two persons A and B perform same amount of work in moving a body through a certain distance d with application of forces acting at angle $45^{\circ}$ and $60^{\circ}$ with the direction of displacement respectively. The ratio of force applied by person A to the force applied by person B is $\frac{1}{\sqrt{x}}$ . The value of x is .....
Correct Answer: 2
Explanation:
Ans. (2)
Given $W_{A} = W_{B}$
$F _ {A} d \cos 4 5 ^ {\circ} = F _ {B} d \cos 6 0 ^ {\circ}$
$F _ {A} \times \frac {1}{\sqrt {2}} = F _ {B} \times \frac {1}{2}$
$\frac {F _ {A}}{F _ {B}} = \frac {\sqrt {2}}{2} = \frac {1}{\sqrt {2}}$
x = 2
Given $W_{A} = W_{B}$
$F _ {A} d \cos 4 5 ^ {\circ} = F _ {B} d \cos 6 0 ^ {\circ}$
$F _ {A} \times \frac {1}{\sqrt {2}} = F _ {B} \times \frac {1}{2}$
$\frac {F _ {A}}{F _ {B}} = \frac {\sqrt {2}}{2} = \frac {1}{\sqrt {2}}$
x = 2
JEE (Main) 2021
Q71
Allen
JEE Main PYQ
MCQ
A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of $0.8\sqrt{gh}$ . The value of work done by the air-friction is :
A.
-0.68 mgh
B.
mgh
C.
1.64 mgh
D.
0.64 mgh
JEE (Main) 2021
Q72
Allen
JEE Main PYQ
MCQ
Statement I: A cyclist is moving on an unbanked road with a speed of $7 \, kmh^{-1}$ and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve ( $g = 9.8 \, m/s^{2}$ )
Statement II: If the road is banked at an angle of $45^{\circ}$ , cyclist can cross the curve of 2m radius with the speed of $18.5 \, kmh^{-1}$ without slipping.
In the light of the above statements,
choose the correct answer from the options given below.
A.
Statement I is incorrect and statement II is correct
B.
Statement I is correct and statement II is incorrect
C.
Both statement I and statement II are false
D.
Both statement I and statement II are true
JEE (Main) 2021
Q73
Allen
JEE Main PYQ
Numerical
The angular speed of truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck engine during this time is ____. (Assuming the acceleration to be uniform).
Correct Answer: 728
Explanation:
Ans. (728)
We know, $\theta = \left(\frac{\omega_1 + \omega_2}{2}\right)t$
Let number of revolutions be N
$\begin{array}{r l} \therefore & 2 \pi N = 2 \pi \left(\frac {9 0 0 + 2 4 6 0}{6 0 \times 2}\right) \times 2 6 \\ & N = 7 2 8 \end{array}$
We know, $\theta = \left(\frac{\omega_1 + \omega_2}{2}\right)t$
Let number of revolutions be N
$\begin{array}{r l} \therefore & 2 \pi N = 2 \pi \left(\frac {9 0 0 + 2 4 6 0}{6 0 \times 2}\right) \times 2 6 \\ & N = 7 2 8 \end{array}$
JEE (Main) 2021
Q74
Allen
JEE Main PYQ
MCQ
The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30° banked road at maximum possible speed without skidding is ____ × $10^{3}$ kg m/s $^{2}$ . [cos 30° = 0.87, $\mu_{s} = 0.2$ ]
A.
10.2
B.
7.2
C.
12.4
D.
6.96
JEE (Main) 2021
Q75
Allen
JEE Main PYQ
MCQ
A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that $r = \frac{L}{\sqrt{2}}$ . The speed of particle will be:
A.
$\sqrt{rg}$
B.
$\sqrt{2rg}$
C.
$2\sqrt{rg}$
D.
$\sqrt{\frac{rg}{2}}$
