Question

Two bodies of masses m, and my are connected by
a light string which passes over a frictionless,
massless pulley. If the pulley is moving upward with
g
uniform acceleration then tension in the string will
2
be​

Answer 1

correct answer is A.

now,

M(

2

3g

)−T=Ma--------(1)

T−m(

2

3g

)=ma---------(2)

Adding equation(1)and(2)we get,

(

2

3g

)(M−m)=a(M+m)

⇒a=

(M+m)

(

2

3g

)(M−m)

so,

(2)⇒T=m(

2

3g

)+ma

⇒T=

(M+m)

m(

2

3g

)+(m)(

2

3g

)(M−m)

⇒T=

(M+m)

3mgM

solution

Answered By

toppr

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Answer 2

Answer:

correct answer is A.

now,

M(

2

3g

)−T=Ma--------(1)

T−m(

2

3g

)=ma---------(2)

Adding equation(1)and(2)we get,

(

2

3g

)(M−m)=a(M+m)

⇒a=

(M+m)

(

2

3g

)(M−m)

so,

(2)⇒T=m(

2

3g

)+ma

⇒T=

(M+m)

m(

2

3g

)+(m)(

2

3g

)(M−m)

⇒T=

(M+m)

3mgM

solution

Explanation:

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