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NEET
Exam
NEET
Subject
Physics
Chapter
Motion in Two Dimensions

NEET · Physics · Motion in Two Dimensions

Given below are two statements. One is labelled as Assertion A and...

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Two identical balls AandA andBthrownwithsamevelocity′thrown with same velocity 'u′attwodifferentangleswithhorizontalattainedthesamerange' at two different angles with horizontal attained the same rangeR.IfAand. If A andBreachedthemaximumheightreached the maximum height{h}_{1}andand{h}_{2}respectively,thenrespectively, thenR = 4\sqrt{{h}_{1}{h}_{2}}<br/><strong>ReasonR:</strong>Productofsaidheights.<br/><strong>Reason R:</strong> Product of said heights.{h}_{1}{h}_{2} = \left(\dfrac{{u}^{2}\sin^{2}\theta}{2g}\right) \cdot \left(\dfrac{{u}^{2}\cos^{2}\theta}{2g}\right)

Options

  1. ABoth A and R are true and R is the correct explanation of A.
  2. BBoth A and R are true but R is NOT the correct explanation of A.
  3. CA is true but R is false.
  4. DA is false but R is true.

Correct answer

A. Both A and R are true and R is the correct explanation of A.

Step-by-step solution

Range of two balls are R1=u2sin⁡2θ1g=R2=u2sin⁡2θ2g{R}_{1} = \dfrac{{u}^{2}\sin 2{\theta}_{1}}{g} = {R}_{2} = \dfrac{{u}^{2}\sin 2{\theta}_{2}}{g}.
Now, maximum height by both balls are h1=u2sin⁡2θ12g{h}_{1} = \dfrac{{u}^{2}\sin^{2}{\theta}_{1}}{2g} and h2=u2cos⁡2θ22g{h}_{2} = \dfrac{{u}^{2}\cos^{2}{\theta}_{2}}{2g}
∴h1h2=(u2sin⁡2θ2g)(u2cos⁡2θ2g)=u2sinθcosθ2g∴ \sqrt{{h}_{1}{h}_{2}} = \left(\dfrac{{u}^{2}\sin^{2}\theta}{2g}\right)\left(\dfrac{{u}^{2}\cos^{2}\theta}{2g}\right) = \dfrac{{u}^{2}sin\theta cos\theta}{2g}=R4= \dfrac{R}{4}
⇒R=4h1h2{⇒ R = 4\sqrt{{h}_{1}{h}_{2}}}^{}
Hence, both A and R are true and R is the correct explanation of A.

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