Question

1. A circular disc with a groove along its diameter is placed
horizontally. A block of mass 1kg is placed as shown. The
co-efficient of friction between the block and all surfaces of
groove in contact is u=2/5. The disc has an acceleration of
25 m/s. Find the acceleration of the block with respect to
disc
[IIT-JEE 2006]
(a) 10 m/s2
(b) 5 m/s a=25m/s2
(c) 20 m/s2
(d) 1 m/s2
cos =4/5 sino=3/5
A
inclined plane making​

Answer 1

Answer:

Answer :

A::B

Solution :

Applying pseudo force ma and resolving it.

Applying F net=mar

macosθ−(f1+f2)=mar

macosθ−μN1−μN2=mar

macosθ−μmasinθ−μmg=mar

⇒ar=acosθ−μasinθ−μg

=25×45−25×25×35−25×10=10m/s2

Explanation:

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