Home
Class 11
PHYSICS
In the figure m(A) = m(B) = m(C) = 60 kg...

In the figure `m_(A) = m_(B) = m_(C) = 60 kg`. The coefficient of friction between `C` and ground is `0.5 B` and ground is `0.3, A` and `B` is `0.4. C` is pulling the string with the maximum possible force without moving. Then the tension in the string connected to `A` will be

A

`120N`

B

`60N`

C

`100N`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
D
Maximum frictional force between `C` and ground `=300Nt`
Max. frictional force between `B` and ground `=360Nt`
So man is unable to pull `B` Hence `T=0`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • DAILY PRACTICE PROBLEMS

    RESONANCE|Exercise dpp 59|5 Videos

  • DAILY PRACTICE PROBLEMS

    RESONANCE|Exercise dpp 60|5 Videos

  • DAILY PRACTICE PROBLEMS

    RESONANCE|Exercise dpp 57|7 Videos

  • CURRENT ELECTRICITY

    RESONANCE|Exercise Exercise|54 Videos

  • ELASTICITY AND VISCOCITY

    RESONANCE|Exercise Advanced Level Problems|9 Videos

Similar Questions

Explore conceptually related problems

In the situation shown coefficient of friction between A and B is 0.5 and between B and C is 0.3 . Friction acting between B and C is xN then (9x)/7 is

In the figure shown coefficient of friction between block A and ground is 0.4 and that between wedge and block B is zero. Wedge is fixed. Find tension is string and acceleration of block A if F = 28 N.

Knowledge Check

  • In the figure shown m_(1)=5 kg,m_(2) = 10 kg & friction coefficient between m_(1) & m_(2) is mu=0.1 and ground is frictionless then :

    A
    If a horizontal force `F=20 N` is applied on `m_(1)` then the friction force acting on `m_(2)` is 5 N in the direction of F
    B
    Maximum amount of horizontal force that can be applied to `m_(2)` such that there is no relative motion between blocks is 15 N
    C
    If a horizontal force `F = 20 N` is applied on `m_(2)` then friction force acting on `m_(1)` is 20/3 N in the direction of applied force
    D
    Maximum amount of horizontal force that can be applied to `m_(1)` such that there is no relative motion between blocks is 8 N

  • In the figure, m_(1)=m_(2)=10 kg . The coefficients of friction between A, B and B , surface are 0.2 . Find the maximum value of m_(3) so that no block slips ( All the pullies are ideal and strings are massless).

    A
    `16 kg`
    B
    `10 kg`
    C
    `18 kg`
    D
    `14 kg`

  • M_(A) = 3 kg ,M_(B) = 4 kg ,M_(C) = 8 kg . Coefficient of friction between any two surface is 0.25 pulley is frictionless and string is massless .A is connected to wall through a massless right rod.

    A
    value of `F` to keep `C` moving with constant speed is `80 N`
    B
    value of `F` to keep `C` moving with constant speed is `120 N`
    C
    if `F is 200 N` then acceleration of `B` is `10 m//s^(-2)`
    D
    to slide `C` towards left ,`F` should be at least ` 50 N` (Take `g = 10m//s^(-2))`

    • Similar Questions

    Explore conceptually related problems

    In the figure , masses m_(1) , m_(2) and M are 20 kg , 5 kg and 50 kg respectively . The coefficient of friction between M and ground is zero . The coefficient of friction between m_(1) and M and that between m_(2) and ground is 0.4 . The pulleys and the string are massless . The string is perfectly horizontal between P_(1) and m_(1) and also between P_(2) and m_(2) . The string is perfectly vertical between P_(1) and P_(2) . An external horizonal force F is applied to the mass M . (Take g = 10 m //s^(2)) (a) Draw a free body diagram for mass M , clearly showing all the forces . (b) Let the magnitude of the force of friction between m_(1) and M be f_(1) and that between m_(2) and ground be f_(2) . For a particular F it is found that f_(1) = 2f_(2) . Find f_(1) and f_(2) . Write down equation of motion of all the masses . Find F , tension in the string and acceleration of the masses .

    A 50 kg man is standing at the centre of a 30 kg platform A. Length of the platform is 10 m and coefficient of friction between the platform and the horizontal ground is 0.2. Man is holding one end of a light rope which is connected to a 50 kg box B. The coefficient of friction between the box and the ground is 0.5. The man pulls the rope so as to slowly move the box and ensuring that he himself does not move relative to the ground. If the shoes of the man does not slip on the platform, calculate how much time it will take for the man to fall off the platform. Assume that rope remains horizontal, and coefficient of friction between shoes and the platform is 0.6.

    Two block A and B of mass m_(A)= 10 kg and m_(A) = 20 kg are place on rough horizontal surface . The blocks are connected with a string. If the coefficient of fraction between block A and ground is mu _(A) = 0.9 and between block B and ground is mu _(B) = 0.3 and the tension in the string in situation as shown in fig force 120 N and 100 N start acting when the system is at rest?

    The coefficient of friction between 4 kg and 5 kg blocks is 0.2 and between 5 kg block and ground is 0.1 respectively. Choose the correct statement : (Take g=10m//s^(2) )

    Two masses M and m are connected by a weightless string. They are pulled by a force F on a frictionless horizontal surface. The tension in the string will be

    Home
    Profile