In the arrangement shown, end $A$ of light inextensible string is pulled up with constant velocity $v$ . The velocity of block $B$ is

\frac{v}{2}

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v

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\frac{v}{3}

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3 v

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Solution

The correct option is C $\frac{v}{3}$
let $V_{B}$ be the velocity of the block $B$

From constraint relation,
$T_{A} . V_{A} = T_{B} . V_{B}$
From diagram, $(T_{B} = 3 T)$
$\Rightarrow T v = 3 T \times V_{B}$ , $V_{B} = \frac{v}{3}$
The correct answer is (c)

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In the arrangement shown, end A of light inextensible string is pulled up with constant velocity v. The velocity of block B is

The correct option is C v3let VB be the velocity of the block B From constraint relation, TA.VA=TB.VB From diagram, (TB=3T) ⇒Tv=3T×VB, VB=v3 The correct answer is (c)

In the arrangement shown, end $A$ of light inextensible string is pulled up with constant velocity $v$ . The velocity of block $B$ is

The correct option is C $\frac{v}{3}$
let $V_{B}$ be the velocity of the block $B$

From constraint relation,
$T_{A} . V_{A} = T_{B} . V_{B}$
From diagram, $(T_{B} = 3 T)$
$\Rightarrow T v = 3 T \times V_{B}$ , $V_{B} = \frac{v}{3}$
The correct answer is (c)