Work, Power & Energy
227 Questions
Start Allen Test
JEE (Main) 2023
Q1
Allen
10. JEE Main PYQ
MCQ
A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s, then the ratio of extension in the spring to its natural length will be:
A.
1 : 2
B.
1 : 1
C.
2 : 3
D.
2 : 5
JEE (Main) 2023
Q2
Allen
10. JEE Main PYQ
Numerical
A car is moving on a circular path of radius 600 m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km/hr is $t(1 - e^{-\pi/2})$ s. The value of t is ____.
Correct Answer: 40
Explanation:
Ans. (40)
$v \frac {d v}{d x} = \frac {v ^ {2}}{R} \Rightarrow \int_ {1 5} ^ {v} \frac {d v}{v} = \frac {1}{R} \int_ {0} ^ {x} d x$
$v = 1 5 e ^ {x / R}$
$\frac {d x}{d t} = 1 5 e ^ {x / R}$
$\int_ {0} ^ {\frac {\pi x}{2}} e ^ {- x / R} d x = 1 5 \int_ {0} ^ {t _ {0}} d t$
$t _ {0} = 4 0 \left(1 - e ^ {- \pi / 2}\right)$
$v \frac {d v}{d x} = \frac {v ^ {2}}{R} \Rightarrow \int_ {1 5} ^ {v} \frac {d v}{v} = \frac {1}{R} \int_ {0} ^ {x} d x$
$v = 1 5 e ^ {x / R}$
$\frac {d x}{d t} = 1 5 e ^ {x / R}$
$\int_ {0} ^ {\frac {\pi x}{2}} e ^ {- x / R} d x = 1 5 \int_ {0} ^ {t _ {0}} d t$
$t _ {0} = 4 0 \left(1 - e ^ {- \pi / 2}\right)$
JEE (Main) 2023
Q3
Allen
10. JEE Main PYQ
Numerical
A body of mass 1 kg begins to move under the action of a time dependent force $\vec{F} = (t\hat{i} + 3t^2\hat{j})N$ . where $\hat{i}$ and $\hat{j}$ are the unit vectors along $x$ and $y$ axis. The power developed by above force, at the time $t = 2s$ . will be ____W.
Correct Answer: 100
Explanation:
Ans. (100)
$\vec {F} = t \hat {i} + 3 t ^ {2} \hat {j}$
$\frac {m d \vec {v}}{d t} = t \hat {i} + 3 t ^ {2} \hat {j}$
$m = 1 k g, \int_ {0} ^ {\vec {v}} d v = \int_ {0} ^ {t} t d t \hat {i} + \int_ {0} ^ {t} 3 t ^ {2} d t \hat {j}$
$\vec {v} = \frac {t ^ {2}}{2} \hat {i} + t ^ {3} \hat {j}$
$\mathrm{Power} = \vec {F} \cdot \vec {V} = \frac {t ^ {3}}{2} + 3 t ^ {5}$
$\mathrm{At} t = 2, \mathrm{power} = \frac {8}{2} + 3 \times 3 2 = 1 0 0$
$\vec {F} = t \hat {i} + 3 t ^ {2} \hat {j}$
$\frac {m d \vec {v}}{d t} = t \hat {i} + 3 t ^ {2} \hat {j}$
$m = 1 k g, \int_ {0} ^ {\vec {v}} d v = \int_ {0} ^ {t} t d t \hat {i} + \int_ {0} ^ {t} 3 t ^ {2} d t \hat {j}$
$\vec {v} = \frac {t ^ {2}}{2} \hat {i} + t ^ {3} \hat {j}$
$\mathrm{Power} = \vec {F} \cdot \vec {V} = \frac {t ^ {3}}{2} + 3 t ^ {5}$
$\mathrm{At} t = 2, \mathrm{power} = \frac {8}{2} + 3 \times 3 2 = 1 0 0$
JEE (Main) 2023
Q4
Allen
10. JEE Main PYQ
MCQ
Identify the correct statements from the following:
(A) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is negative.
(B) Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative.
(C) Work done by friction on a body sliding down an inclined plane is positive.
(D) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity in zero.
(E) Work done by the air resistance on an oscillating pendulum in negative.
Choose the correct answer from the options given below:
(A) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is negative.
(B) Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative.
(C) Work done by friction on a body sliding down an inclined plane is positive.
(D) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity in zero.
(E) Work done by the air resistance on an oscillating pendulum in negative.
Choose the correct answer from the options given below:
A.
B and E only
B.
A and C only
C.
B, D and E only
D.
B and D only
2022
Q5
Allen
10. JEE Main PYQ
MCQ
A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is $\frac{K}{\pi}$ rev./min. The value of K is:
(Assume the string is massless and unstretchable)
(JEE (Main) 2022)
A.
400
B.
300
C.
600
D.
800
JEE (Main) 2022
Q6
Allen
10. JEE Main PYQ
MCQ
Potential energy as a function of r is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$ , where r is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be:
A.
$\left(\frac{A}{B}\right)^{\frac{1}{5}}$
B.
$\left(\frac{B}{A}\right)^{\frac{1}{5}}$
C.
$\left(\frac{2A}{B}\right)^{\frac{1}{5}}$
D.
$\left(\frac{B}{2A}\right)^{\frac{1}{5}}$
JEE (Main) 2022
Q7
Allen
10. JEE Main PYQ
MCQ
A particle experiences a variable force $\vec{F} = (4x\hat{\imath} + 3y^2\hat{j})$ in a horizontal $x-y$ plane. Assume distance in meters and force in newton. If the particle moves from point (1, 2) to point (2, 3) in the $x-y$ plane, the Kinetic Energy changes by
A.
$50.0 J$
B.
$12.5 J$
C.
$25.0 J$
D.
$0 J$
JEE (Main) 2022
Q8
Allen
10. JEE Main PYQ
Numerical
A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\frac{h}{2}$ . The spring constant is ____ $Nm^{-1}$ . (Use $g = 10 ms^{-2}$ )
Correct Answer: 120
Explanation:
Ans. (120)
By energy conservation
$P E = K E$
$m g \left(H + \frac {H}{2}\right) = \frac {1}{2} k x ^ {2} \left(x = \frac {H}{2}\right)$
$0.1 0 0 \times 1 0 \times \frac {3}{2} (0.1 0) = \frac {1}{2} k (0.0 5 \times 0.0 5)$
$k = \frac {3 \times 0.1 0}{0.0 5 \times 0.0 5} = \frac {3 \times 1 0 0 0}{2 5} = 1 2 0 N / m$
By energy conservation
$P E = K E$
$m g \left(H + \frac {H}{2}\right) = \frac {1}{2} k x ^ {2} \left(x = \frac {H}{2}\right)$
$0.1 0 0 \times 1 0 \times \frac {3}{2} (0.1 0) = \frac {1}{2} k (0.0 5 \times 0.0 5)$
$k = \frac {3 \times 0.1 0}{0.0 5 \times 0.0 5} = \frac {3 \times 1 0 0 0}{2 5} = 1 2 0 N / m$
JEE (Main) 2021
Q9
Allen
10. 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
Q10
Allen
10. 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
Q11
Allen
10. 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/sĀ²)
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
Q12
Allen
10. 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
Q13
Allen
10. 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
Q14
Allen
10. 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
Q15
Allen
10. 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
Q16
Allen
10. 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
Q17
Allen
10. 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
Q18
Allen
10. 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
Q19
Allen
10. 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
Q20
Allen
10. 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
Q21
Allen
10. JEE Main PYQ
MCQ
The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30Ā° banked road at maximum possible speed without skidding is ____ Ć $10^{3}$ kg m/s $^{2}$ . [cos 30Ā° = 0.87, $\mu_{s} = 0.2$ ]
A.
10.2
B.
7.2
C.
12.4
D.
6.96
JEE (Main) 2021
Q22
Allen
10. 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.
$\sqrt{\frac{rg}{2}}$
JEE (Main) 2021
Q23
Allen
10. JEE Main PYQ
MCQ
A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second?
Given: $1 \, ly = 9.46 \times 10^{15} \, m$ $1 \, AU = 1.5 \times 10^{11} \, m$
A.
$4.1 \times 10^{8} \, s$
B.
$4.5 \times 10^{10} \, s$
C.
$3.5 \times 10^{6} \, s$
D.
$7.2 \times 10^{8} \, s$
JEE (Main) 2020
Q24
Allen
10. JEE Main PYQ
MCQ
A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to:
(1 HP = 746 W, g = 10 ms $^{-2}$ )
A.
1.7 ms $^{-1}$
B.
2.0 ms $^{-1}$
C.
1.9 ms $^{-1}$
D.
1.5 ms $^{-1}$
JEE (Advanced) 2020
Q25
Allen
12. JEE Advanced PYQ
MSQ
A student skates up a ramp that makes an angle $30^{\circ}$ with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed $v_{0}$ and wants to turn around over a semicircular path xyz of radius R during which he/she reaches a maximum height h (at point y) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then (g is the acceleration due to gravity)
A.
$v_{0}^{2} - 2gh = \frac{1}{2} gR$
B.
$v_{0}^{2} - 2gh = \frac{\sqrt{3}}{2} gR$
C.
the centripetal force required at points x and z is zero
D.
the centripetal force required is maximum at points x and z