(C) The time taken by a coin to reach the floor is given by the kinematic equation $h = \frac{1}{2} gt^2$ (since initial velocity $u = 0$).Thus,$t = \

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The time taken by a coin to reach the floor is given by the kinematic equation $h = \frac{1}{2} gt^2$ (since initial velocity $u = 0$).Thus,$t = \sqrt{\frac{2h}{g}}$.In a stationary lift,the effective acceleration is $g$,so $t_1 = \sqrt{\frac{2h}{g}}$.In a lift moving upward with constant acceleration $a$,the effective acceleration experienced by the coin is $g' = g + a$.Therefore,$t_2 = \sqrt{\frac{2h}{g+a}}$.Since $g' = g + a > g$,it follows that the denominator in the expression for $t_2$ is larger than that for $t_1$.Thus,$t_2 < t_1$.

Class 11 Physics Laws of Motion Apparent weight and Pseudo Force

Medium

$A$ coin is dropped in a lift. It takes time $t_1$ to reach the floor when the lift is stationary. It takes time $t_2$ when the lift is moving up with constant acceleration $a$. Then:

A
$t_1 = t_2$
B
$t_1 < t_2$
C
$t_2 < t_1$
D
$t_1 > t_2$ is incorrect,$t_2 > t_1$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Laws of Motion

Subtopic

Apparent weight and Pseudo Force

$A$ car is moving with a constant speed of $20\,m/s$ on a circular horizontal track of radius $40\,m$. $A$ bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be: (Take $g = 10\,m/s^2$)

MediumJEE MAIN 2023

View Solution

$A$ block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ($g$ is acceleration due to gravity) will be

DifficultAP EAMCET 2020

View Solution

$A$ person of $60 \,kg$ mass is in a lift which is coming down such that the man exerts a force of $150 \,N$ on the floor of the lift. Then the acceleration of the lift is $\left(g=10 \,ms^{-2}\right)$ (in $\,ms^{-2}$)

EasyAP EAMCET 2017

View Solution

$A$ person in an elevator accelerating upwards with an acceleration of $2\,ms^{-2}$,tosses a coin vertically upwards with a speed of $20\,ms^{-1}$. After how much time will the coin fall back into his hand? (Take $g = 10\,ms^{-2}$)

Medium

View Solution

An elevator car whose floor-to-ceiling distance is $2.7\,m$ starts ascending with a constant acceleration of $1.2\,m/s^2$. After $2\,s$ of the start,a bolt falls from the ceiling of the car. The free fall time of the bolt is $(g=9.8\,m/s^2)$.

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ coin is dropped in a lift. It takes time $t_1$ to reach the floor when the lift is stationary. It takes time $t_2$ when the lift is moving up with constant acceleration $a$. Then:

Similar Questions

$A$ car is moving with a constant speed of $20\,m/s$ on a circular horizontal track of radius $40\,m$. $A$ bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be: (Take $g = 10\,m/s^2$)

$A$ block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ($g$ is acceleration due to gravity) will be

$A$ person of $60 \,kg$ mass is in a lift which is coming down such that the man exerts a force of $150 \,N$ on the floor of the lift. Then the acceleration of the lift is $\left(g=10 \,ms^{-2}\right)$ (in $\,ms^{-2}$)

$A$ person in an elevator accelerating upwards with an acceleration of $2\,ms^{-2}$,tosses a coin vertically upwards with a speed of $20\,ms^{-1}$. After how much time will the coin fall back into his hand? (Take $g = 10\,ms^{-2}$)

An elevator car whose floor-to-ceiling distance is $2.7\,m$ starts ascending with a constant acceleration of $1.2\,m/s^2$. After $2\,s$ of the start,a bolt falls from the ceiling of the car. The free fall time of the bolt is $(g=9.8\,m/s^2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module