A bomb projected from ground at an angle θθ≠90° with horizontal explodes into two fragments of equal mass at top most point of its trajectory. If one of the fragment returns to point of projection then ratio of de-Broglie wavelength of second fragment just after explosion to bomb just before explosion is :
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answer is 0000.67.
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Detailed Solution
From the law of conservation of momentummucosθ=−m2ucosθ+m2v1v1=3ucosθ…(1)λD=hmv…(2)The ratio of De-Broglie wavelength⇒λDλD2=hm23ucosθhmucosθ=23Therefore, the correct answer is 0.67.