A bob of mass m is suspended from point O by a massless string of length l as shown- At the bottommost point it is given a velocity u- -12gl for l-1m and m-1 kg- match the following two columns when string becomes horizontal g- 10ms-2

Given : #160; u = √(12 gl) #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; m = 1 kg #160; #160; ...

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Byju's Answer

Standard XII

Physics

Work Energy Theorem Application

A bob of mass...

Question

A bob of mass $m$ is suspended from point $O$ by a massless string of length $l$ as shown. At the bottommost point it is given a velocity $u = \sqrt{12 g l}$ for $l = 1 m$ and $m = 1 k g$ , match the following two columns when string becomes horizontal $(g = 10 m s^{- 2})$

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Solution

Given : $u = \sqrt{12 g l}$ $m = 1 k g$ $l = 1$ m $g = 10 m / s^{2}$

Let the velocity of the bob be $v$ when the string becomes horizontal.
Using work-energy theorem : $W = Δ K . E$
$∴$ $- m g l = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

OR $- 2 g l = v^{2} - (12 g l)$ $⟹ v = \sqrt{10 g l}$

$∴$ $v = \sqrt{10 \times 10 \times 1} = 10$ m/s

Radial acceleration, $a_{r} = \frac{v^{2}}{l} = \frac{10^{2}}{1} = 100$ $m / s^{2}$
Tension in the string, $T = m a_{r} = 1 \times 100 = 100$ $N$
Tangential acceleration, $a_{t} = m g = 1 \times 10 = 10$ $N$

$∴$ Total acceleration, $a = \sqrt{a_{r}^{2} + a_{t}^{2}} = \sqrt{(100)^{2} + (10)^{2}} = 10 \sqrt{101} = 100.5$ $m / s^{2}$

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A bob of mass m is suspended from point O by a massless string of length l as shown. At the bottommost point it is given a velocity u=√12gl for l=1m and m=1kg, match the following two columns when string becomes horizontal (g=10ms−2)

A bob of mass $m$ is suspended from point $O$ by a massless string of length $l$ as shown. At the bottommost point it is given a velocity $u = \sqrt{12 g l}$ for $l = 1 m$ and $m = 1 k g$ , match the following two columns when string becomes horizontal $(g = 10 m s^{- 2})$