Home
Class 12
PHYSICS
Consider the case of two bodies of masse...

Consider the case of two bodies of masses `m_(1)` and `m_(2)` which are connected by light inextensible string passing over a light smooth pulley as shown in the figure. The expression for acceleration of the system and tension of the string are expressed below under different situations :
(`i`) when `m_(1) gt m_(2)`. In this case
`a=((m_(1)-m_(2))/(m_(1)+m_(2)))g` and `T=((2m_(1)m_(2))/(m_(1)+m_(2)))g`
(`ii`) when `m_(2) gt m_(1)`. In this case
`a=((m_(2)-m_(1))/(m_(1)+m_(2)))g` and `T=((2m_(1)m_(2))/(m_(1)+m_(2)))g`
(`iii`) When `m_(1)=m_(2)=m`. In this case `a=0`, `T=mg`
If `m_(1)=10kg`, `m_(2)=6kg` and `g=10m//s^(2)`
If the pulley is pulled upward with acceleration equal to the acceleration due to gravity, what will be the tension in the string

A

75N

B

150N

C

300N

D

500N

Text Solution

Verified by Experts

The correct Answer is:
B
`2m_(1)g-T.=m_(1)a. " " ...(1) `
`T.-2m_(2)g=m_(2)a. " "...(2)`
On solving (1) and (2)
`a.=(g)/(2)=5m//s^(2)`
`T.=150N`
Promotional Banner

Similar Questions

Explore conceptually related problems

Consider the case of two bodies of masses m_(1) and m_(2) which are connected by light inextensible string passing over a light smooth pulley as shown in the figure. The expression for acceleration of the system and tension of the string are expressed below under different situations : ( i ) when m_(1) gt m_(2) . In this case a=((m_(1)-m_(2))/(m_(1)+m_(2)))g and T=((2m_(1)m_(2))/(m_(1)+m_(2)))g ( ii ) when m_(2) gt m_(1) . In this case a=((m_(2)-m_(1))/(m_(1)+m_(2)))g and T=((2m_(1)m_(2))/(m_(1)+m_(2)))g ( iii ) When m_(1)=m_(2)=m . In this case a=0 , T=mg If m_(1)=10kg , m_(2)=6kg and g=10m//s^(2) What is the tension in the string ?

Consider the case of two bodies of masses m_(1) and m_(2) which are connected by light inextensible string passing over a light smooth pulley as shown in the figure. The expression for acceleration of the system and tension of the string are expressed below under different situations : ( i ) when m_(1) gt m_(2) . In this case a=((m_(1)-m_(2))/(m_(1)+m_(2)))g and T=((2m_(1)m_(2))/(m_(1)+m_(2)))g ( ii ) when m_(2) gt m_(1) . In this case a=((m_(2)-m_(1))/(m_(1)+m_(2)))g and T=((2m_(1)m_(2))/(m_(1)+m_(2)))g ( iii ) When m_(1)=m_(2)=m . In this case a=0 , T=mg If m_(1)=10kg , m_(2)=6kg and g=10m//s^(2) then what is the acceleration of masses ?

Two bodies of masses m_(1) " and " m_(2) are connected a light string which passes over a frictionless massless pulley. If the pulley is moving upward with uniform acceleration (g)/(2) , then tension in the string will be

Three blocks of masses m_(1)=4kg, m_(2)=2kg, m_(3)=4kg are connected with ideal strings passing over a smooth, massless pulley as shown in figure. The acceleration of blocks will be (g=10m//s^(2))

Two masses m and M(m lt M) are joined by a light string passing over a smooth and light pulley (as shown)

Two blocks of masses m_(1) and m_(2)(m_(1) gt m_(2)) connected by a massless string passing over a frictionless pulley Magnitude of acceleration of blocks would be

Two masses m_(1) and m_(2) are connected to the end of a string passing over a smooth pulley . The magnitude of tension in the string is T and the masses are moving with magnitude of acceleration a . If the masses are interchanged , then

Blocks of mass M_(1) and M_(2) are connected by a cord which passes over the pulleys P_(1) and P_(2) as shown in the figure. If there is no friction, the acceleration of the block of mass M_(2) will be

Two bodies with masses m_1 and m_2(m_1gtm_2) are joined by a string passing over fixed pulley. Assume masses of the pulley and thread negligible. Then the acceleration of the centre of mass of the system (m_1+m_2) is

Two bodies of masses m_(1)(m_(2) gt m_(1)) are connected by a light inextensible string which passes through a smooth fixed pulley. What is the instantaneous power delivered by an external agent to pull m_(1) with constant velocity vec(V) ?