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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Motion in Two Dimensions

NTA Abhyas 2020 · Physics · Motion in Two Dimensions

An inclined plane is located at angle α = 53 ° to the horizontal....

An inclined plane is located at angle α=53°\alpha = 53 ° to the horizontal. There is a hole at point BB in the inclined plane as shown in the figure. A particle is projected along the plane with speed v0{v}_{0} at an angle β=37°\beta = 37 ° to the horizontal in such a way so that it gets into the hole. Neglect any type of friction. Find the speed v0{v}_{0} (in ms−1ms^{- 1})ifh=1mif h = 1m and l=8ml = 8m.

An inclined plane is located at angle α = 53 ° to the horizontal. There is a hole at point B in the inclined plane as

Options

  1. A9
  2. B3
  3. C6
  4. D12

Correct answer

A. 9

Step-by-step solution

An inclined plane is located at angle α = 53 ° to the horizontal. There is a hole at point B in the inclined plane as

h=ltanβ−12gsin\alphal2v02cos⁡2βh = ltan \beta - \dfrac{1}{2}\dfrac{gsin \alphal^{2}}{{v}_{0}^{2}\cos^{2} \beta}
v0=gsin\alphal22cos⁡2β(ltanβ−h){v}_{0} = \sqrt{\dfrac{gsin \alphal^{2}}{2\cos^{2} \beta\left(ltan \beta - h\right)}}
=10×45×8×82×45×45(8×34−1)=9ms−1= \sqrt{\dfrac{10 \times \dfrac{4}{5} \times 8 \times 8}{2 \times \dfrac{4}{5} \times \dfrac{4}{5}\left(8 \times \dfrac{3}{4} - 1\right)}} = 9ms^{- 1}
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