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Standard XII

Physics

SHM in Vertical Spring Block

A bob weighin...

Question

A bob weighing $50$ gram hangs vertically at the end of a string $50 c m$ long. If $20$ gram force is applied horizontally, by how much distance the bob is pulled aside from its initial position when it reaches it equilibrium position?

19.57 c m

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13.87 c m

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17.57 c m

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27.57 c m

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Solution

The correct option is C $17.57 c m$
Here, $m = 50$ gram; $F = 20 g f$ . Let the bob be in equilibrium when it is at location $B$ Fig.2(c).17. Then
$\frac{F}{C B} = \frac{m g}{O C} = \frac{T}{B O}$
or $\frac{C B}{O C} = \frac{F}{m g} = \frac{20 \times 980}{50 \times 980} = 0.4$ $tan θ = \frac{C B}{O C} = 0.4 = tan 21^{o} 48$
so $θ = 21^{o} 48$
In $△ O C B, sin 21^{o} 48 = \frac{C B}{O B}$
or $C B = O B sin 21^{o} 48$
$= 50 \times 0.3714 = 17.57 c m$

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A bob weighing 50 gram hangs vertically at the end of a string 50 cm long. If 20 gram force is applied horizontally, by how much distance the bob is pulled aside from its initial position when it reaches it equilibrium position?

The correct option is C 17.57 cmHere, m=50 gram; F=20 gf. Let the bob be in equilibrium when it is at location B Fig.2(c).17. ThenFCB=mgOC=TBO or CBOC=Fmg=20×98050×980=0.4 tanθ=CBOC=0.4=tan21o48 so θ=21o48 In △OCB,sin21o48=CBOB or CB=OBsin21o48 =50×0.3714=17.57 cm

A bob weighing $50$ gram hangs vertically at the end of a string $50 c m$ long. If $20$ gram force is applied horizontally, by how much distance the bob is pulled aside from its initial position when it reaches it equilibrium position?