A ball is released from rest from the top of a tower

Answer

Verified

\[\begin{align} & {{s}_{1}}=0\times 1+\dfrac{1}{2}g{{(1)}^{2}} \\ & {{s}_{1}}=\dfrac{g}{2} \\ \end{align}\]Distance travelled in second second = distance travelled in two second minus distance travelled in first second of motion.\[\begin{align} & {{s}_{2}}=0\times 2+\dfrac{1}{2}g{{(2)}^{2}}-{{s}_{1}} \\ & {{s}_{2}}=\dfrac{4g}{2}-\dfrac{g}{2} \\ & {{s}_{2}}=\dfrac{3g}{2} \\ \end{align}\]Distance travelled in third second = distance travelled in three seconds minus distance travelled in first two second.\[\begin{align} & {{s}_{3}}=0\times 3+\dfrac{1}{2}g{{(3)}^{2}}-0\times 2+\dfrac{1}{2}g{{(2)}^{2}} \\ & {{s}_{3}}=\dfrac{9g}{2}-\dfrac{4g}{2} \\ & {{s}_{3}}=\dfrac{5g}{2} \\ \end{align}\]Now, taking the ratio of work done by force of gravity in first, second and third second of motion:\[\begin{align} & {{W}_{1}}:{{W}_{2}}:{{W}_{3}} \\ & =F{{s}_{1}}:F{{s}_{2}}:F{{s}_{3}} \\ & ={{s}_{1}}:{{s}_{2}}:{{s}_{3}} \\ & =\dfrac{g}{2}:\dfrac{3g}{2}:\dfrac{5g}{2} \\ & =1:3:5 \\ \end{align}\]Hence, the ratio of work done by force of gravity in first, second and third second of motion is 1:3:5Therefore, option C is correct. Note: This question can also be done by applying the formula of distance travelled by body in \[{{n}^{th}}\] second.\[{{s}_{n}}=u+\dfrac{a}{2}(2n-1)\]Just put u=0 and a=g\[{{s}_{1}}=\dfrac{g}{2}\], \[{{s}_{2}}=\dfrac{3g}{2}\], \[{{s}_{3}}=\dfrac{5g}{2}\]If the particle starts from rest and moves with constant acceleration then the ratio of displacement in equal interval of time is.\[{{s}_{1}}:{{s}_{2}}:{{s}_{3}}:{{s}_{4}}...=1:3:5:7...\]

Q: A ball is released from rest from top of a tower. The retardation due to air resistance is bv , where b is 10 per second and velocity v is in ms-1. The velocity of ball at t = 1/10 sec is n/27 m/s . Find the value of n. (Given, e = 2.7)

Ans. 7

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