(B) The acceleration of the box is $a = 10\,m/s^2$. In the non-inertial frame of the box,a pseudo force $F_p = ma$ acts on the body in the direction o

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(B) The acceleration of the box is $a = 10\,m/s^2$. In the non-inertial frame of the box,a pseudo force $F_p = ma$ acts on the body in the direction opposite to the acceleration of the box. Here,$m = 1\,kg$ and $a = 10\,m/s^2$,so the pseudo force $F_p = 1 \times 10 = 10\,N$ acts horizontally. The gravitational force $mg = 1 \times 10 = 10\,N$ acts vertically downwards. In the equilibrium position relative to the experimenter,the tension $T$ in the string must balance the resultant of the gravitational force and the pseudo force. $T = \sqrt{(mg)^2 + (ma)^2} = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}\,N$.

Class 11 Physics Newton's Laws of Motion and Friction Apparent weight and Pseudo Force

Easy

Inside a horizontally moving box,an experimenter finds that when an object is placed on a smooth horizontal table and is released,it moves with an acceleration of $10\,m/s^2$. In this box,if a $1\,kg$ body is suspended with a light string,the tension in the string in the equilibrium position (with respect to the experimenter) will be $.........\,N$. (Take $g = 10\,m/s^2$)

A
$10$
B
$10\sqrt{2}$
C
$20$
D
$0$

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Class 11 Questions

Class 11

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Chapter

Newton's Laws of Motion and Friction

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Apparent weight and Pseudo Force

$A$ person standing on a spring balance inside a stationary lift measures $60 \, kg$. The weight of that person if the lift descends with a uniform downward acceleration of $1.8 \, m/s^{2}$ will be $_{-} N$. $[g = 10 \, m/s^{2}]$

MediumJEE MAIN 2021

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The minimum acceleration that must be imparted to the cart in the figure so that the block $A$ will not fall (given $\mu$ is the coefficient of friction between the surfaces of the block and the cart) is given by

Difficult

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The weight of a man in a lift moving in an upward direction with an acceleration '$a$' is $660 \text{ N}$. When the lift moves in the downward direction with the same acceleration, his weight is found to be $380 \text{ N}$. The real weight of the man when the lift is at rest is: (in $\text{ N}$)

DifficultMHT CET 2026

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$A$ car is moving with a uniform speed on a level road. Inside the car,there is a balloon filled with helium and attached to a piece of string tied to the floor. The string is observed to be vertical. The car now takes a left turn while maintaining the same speed on the level road. The balloon in the car will:

Easy

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$A$ 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

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$A$ person standing on a spring balance inside a stationary lift measures $60 \, kg$. The weight of that person if the lift descends with a uniform downward acceleration of $1.8 \, m/s^{2}$ will be $_{-} N$. $[g = 10 \, m/s^{2}]$

The minimum acceleration that must be imparted to the cart in the figure so that the block $A$ will not fall (given $\mu$ is the coefficient of friction between the surfaces of the block and the cart) is given by

The weight of a man in a lift moving in an upward direction with an acceleration '$a$' is $660 \text{ N}$. When the lift moves in the downward direction with the same acceleration, his weight is found to be $380 \text{ N}$. The real weight of the man when the lift is at rest is: (in $\text{ N}$)

$A$ 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 $g = 10\,m/s^2$)

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