Kewill B
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If incident frequency is doubled new (KE) will be (a) 2* ' (b) 2* - 0 (c) 2k + 0 (d) 2k + 2tf> (e) 2k-2 Solution (c) k=hf-Q or 2k = h(2f) - 2 = 2hf-
Find the stopping potential (a) (O 1.37V 1.57V (b) (d) 1.47 V 1.51V (b) If incident frequency is doubled new (KE) will be (a) 2k (b) 2k - <|> (c) 2k + if) (d) 2k +
V(x) E a2 a 2 b 2 x – b – 2 –V1 (iv) = Ans. It is clear that the total energy of the body In the region x > a, the P.E. (V0 ) has a value greater than E. So, K.E. will be
(b) Work done (frictional force) = −f.s = −1.96 126 = −247J (c) work done (net In the region x > a, the P.E. (V0 ) has a value greater than E. So, K.E. will be
(ii) In is region x < a and x > b, the P.E. greater than the total energy of the is V particle. 0 which So, K.E. will be negative in this region. Thus, the particle cannot
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