JEE Main 2021 — Physics Mechanics

Let the speed of bob at lowest position be ${v}_{1}$v1​ and at the highest position be ${v}_{2}$v2​.

Maximum tension is at lowest position and minimum tension is at the highest position.

Now, using, conservation of mechanical energy,

$$\frac{1}{2}m{v}_{1}^{2}=\frac{1}{2}m{v}_{2}^{2}+mg2l$$21​mv12​=21​mv22​+mg2l

$$\Rightarrow {v}_{1}^{2}={v}_{2}^{2}+4gl....(1)$$⇒v12​=v22​+4gl....(1)

Now, ${T}_{\mathrm{max}}-mg=\frac{m{v}_{1}^{2}}{l}$Tmax​−mg=lmv12​​

$$\Rightarrow {T}_{\mathrm{max}}=mg+\frac{m{v}_{1}^{2}}{l}$$⇒Tmax​=mg+lmv12​​

&{T}_{\mathrm{min}}+mg=\frac{m{v}_{2}^{2}}{l}

$$\Rightarrow {T}_{\mathrm{min}}=\frac{m{v}_{2}^{2}}{l}-mg$$⇒Tmin​=lmv22​​−mg

$$\frac{{T}_{\mathrm{max}}}{{T}_{\mathrm{min}}}=\frac{5}{1}$$Tmin​Tmax​​=15​

$$\Rightarrow \frac{mg+\frac{m{v}_{1}^{2}}{l}}{\frac{m{v}_{2}^{2}}{l}-mg}=\frac{5}{1}$$⇒lmv22​​−mgmg+lmv12​​​=15​

$$\Rightarrow mg+\frac{m{v}_{1}^{2}}{l}=[\frac{m{v}_{2}^{2}}{l}-mg]5$$⇒mg+lmv12​​=[lmv22​​−mg]5

$$\Rightarrow mg+\frac{m}{l}[{v}_{2}^{2}+4gl]=\frac{5m{v}_{2}^{2}}{l}-5mg$$⇒mg+lm​[v22​+4gl]=l5mv22​​−5mg

$$\Rightarrow mg+\frac{m{v}_{2}^{2}}{l}+4mg=\frac{5m{v}_{2}^{2}}{l}-5mg$$⇒mg+lmv22​​+4mg=l5mv22​​−5mg

$$\Rightarrow 10mg=\frac{4m{v}_{2}^{2}}{l}$$⇒10mg=l4mv22​​

$${v}_{2}=\frac{10\times 10\times 1}{4}$$v2​=410×10×1​

Thus, the velocity of the bob at the highest position is $5m{s}^{-1}.$5ms−1.