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QJEEM23Ph0c44: JEE Main 2023 Physics - Class 11 JEEM23Ph0c44 Solution: A body is released from a height equal to the radius of the…

A body is released from a height equal to the radius (R)(\mathrm{R}) of the earth. The velocity of the body when it strikes the surface of the earth will be

(Given g=g= acceleration due to gravity on the earth.)

Multiple Choice Options

  • A gR2\sqrt{\frac{g R}{2}}
  • B 4gR\sqrt{4 g R}
  • C 2gR\sqrt{2 g R}
  • D gR\sqrt{g R}

Step-by-Step Verified Solution

JEE Main 2023 (Online) 15th April Morning Shift Physics - Gravitation Question 53 English Explanation
By conservation of mechanical energy

amp;Ui+Ki=Uf+Kiamp;−GMm2R+0=−GMmR+12mv2amp;GMm2R=12mv2amp;v=GMR=gR\begin{aligned}& \mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{i}} \\\\& -\frac{\mathrm{GMm}}{2 \mathrm{R}}+0=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^2 \\\\& \frac{\mathrm{GMm}}{2 \mathrm{R}}=\frac{1}{2} \mathrm{mv}^2 \\\\& \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}\end{aligned}

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