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
Answers
Answered by
0
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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Answered by
1
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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