In the given arrangement of a doubly inclined plane, two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
In the given arrangement of a doubly inclined plane, two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of m, for which M = 10 kg will move down with an acceleration of 2 $m / s^{2}$ , is (take g = 10 $m / s^{2}$ and tan 37° = 3/4) [2024]

1 6.5 kg

2 4.5 kg

3 2.25 kg

4 9 kg

Ans.
(2)

For M block

$10 g \sin 53 ° - μ (10 g) \cos 53 ° - T = 10 \times 2$

$T = 80 - 15 - 20$

$T = 45 N$

For m block

$T - m g \sin 37 ° - μ m g \cos 37 ° = m \times 2$

$45 = 10 m$

$m = 4.5 kg$

Want Us to Email you About Special Offers & Updates?