Question

A solid cylinder of mass m = 4 kg and radius R = 10 cm has two ropes wrapped around it, one near each end. The cylinder is held horizontally by fixing the two free ends of the cords to the hooks on the ceiling such that both the cords to the hooks on the ceiling such that both the cords are exactly verical. The cylinder is released to fall under gravity. The linear acceleration of the cylinder is n/n+1 g then n = ?​

Answer 1

Answer:

Correct option is

A

6.53N

Let a be the acceleration when the cylinder is falling freely.

m×a=m×g−2×T−−−−−−−−(1)

Let r be the radius of cylinder. If torque acting on both the ends are similar then net torque is given by,

T=T

1

+T

2

=2×r×T=1×α−−−−−−(2)

Where I is moment of inertai and α is angular acceleration due to rod rotation.

if we substitute for I and α in the above equation we get,

2×r×T=

2

1

mr

2

(

r

a

)=

2

1

m×r×a

hence,T=

4

1

m×a

Substituting the above equation for T in eq

n

.(1), we get a=(2/3)g

So,T=

4

1

×4×

3

2

×9.8=6.53

Hence,

option (A) is correct answer.

Explanation:

Plz Mark as brainliest

Give a thnx

Answer 2

Answer:

6.53N

Explanation:

m= 4kg

r = 10cm, 0.1m

a = ?

let us find the moment of inertia of solid cylinder

I = MR²/2

4(0.1)²/2

0.04/2

= 0.02

acceleration is

a=2/3(g)

2/3(9.8)

=6.53/s²

T = a (as it is falling under gravity)

T= 6.53 N