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NEET 2025
Exam
NEET
Year
2025
Subject
Physics
Chapter
Work, Power and Energy

NEET 2025 · Physics · Work, Power and Energy

A bob of heavy mass m is suspended by a light string of length l ....

A bob of heavy mass mm is suspended by a light string of length ll. The bob is given a horizontal velocity v0\mathrm{v}_0 as shown in figure. If the string gets slack at some point P making an angle θ\theta from the horizontal, the ratio of the speed v of the bob at point PP to its initial speed v0v_0 is : A bob of heavy mass m is suspended by a light string of length l . The bob is given a horizontal velocity v_0 as shown

Options

  1. A(sin⁡θ)12(\sin \theta)^{\frac{1}{2}}
  2. B(12+3sin⁡θ)12\left(\frac{1}{2+3 \sin \theta}\right)^{\frac{1}{2}}
  3. C(cos⁡θ2+3sin⁡θ)12\left(\frac{\cos \theta}{2+3 \sin \theta}\right)^{\frac{1}{2}}
  4. D(sin⁡θ2+3sin⁡θ)12\left(\frac{\sin \theta}{2+3 \sin \theta}\right)^{\frac{1}{2}}

Correct answer

D. (sin⁡θ2+3sin⁡θ)12\left(\frac{\sin \theta}{2+3 \sin \theta}\right)^{\frac{1}{2}}

Step-by-step solution

A bob of heavy mass m is suspended by a light string of length l . The bob is given a horizontal velocity v_0 as shown
 C.O.M.E. 12mv02=mgℓ(1+sin⁡θ)+12mvp2....(1) At pt P Tp+mgsin⁡θ=mvp2ℓ( as Tp=0)mgsin⁡θ=mvp2ℓ⇒mvp2=mgℓsin⁡θ…….(ii)\text{ C.O.M.E. } \frac{1}{2} \mathrm{mv}_0^2=\mathrm{mg} \ell(1+\sin \theta)+\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^2....(1) \text{ At pt P } \mathrm{Tp}+\mathrm{mg} \sin \theta=\frac{\mathrm{mv}_{\mathrm{p}}^2}{\ell} \quad\left(\text{ as } \mathrm{T}_{\mathrm{p}}=0\right) \mathrm{mg} \sin \theta=\frac{\mathrm{mv}_{\mathrm{p}}^2}{\ell} \Rightarrow \mathrm{mv}_{\mathrm{p}}^2=\mathrm{mg} \ell \sin \theta \ldots \ldots .(\mathrm{ii}) From (i) & (ii) 12mv02=mgℓ(1+sin⁡θ)+12mgℓsin⁡θv02=2gℓ(1+sin⁡θ)+gℓsin⁡θv0=2gℓ+3gℓsin⁡θ....(iii)\begin{aligned} & \frac{1}{2} m v_0^2=m g \ell(1+\sin \theta)+\frac{1}{2} m g \ell \sin \theta \\ & v_0^2=2 g \ell(1+\sin \theta)+g \ell \sin \theta \\ & v_0=\sqrt{2 g \ell+3 g \ell \sin \theta}....(iii) \end{aligned}vp=gℓsin⁡θv0=2 gℓ+3 gℓsin⁡θvpv0=sin⁡θ2+3sin⁡θ\begin{aligned} & \mathrm{v}_{\mathrm{p}}=\sqrt{\mathrm{g} \ell \sin \theta} \\ & \mathrm{v}_0=\sqrt{2 \mathrm{~g} \ell+3 \mathrm{~g} \ell \sin \theta} \\ & \frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_0}=\sqrt{\frac{\sin \theta}{2+3 \sin \theta}} \end{aligned}$
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