Work Energy And Power Question 43
Question: If the K.E. of a body is increased by 300%, its momentum will increase by [JIPMER 1978; AFMC 1993; RPET 1999; CBSE PMT 2002]
Options:
A) 100%
B) 150%
C) $ \sqrt{300}% $
D) 175%
Show Answer
Answer:
Correct Answer: A
Solution:
Let initial kinetic energy, $ E_1=E $
Final kinetic energy, $ E_2=E+300% $ of E = 4E As $ P\propto \sqrt{E} $
therefore $ \frac{P_2}{P_1}=\sqrt{\frac{E_2}{E_1}}=\sqrt{\frac{4E}{E}}=2 $
therefore $ P_2=2P_1 $
therefore $ P_2=P_1+100% $ of $ P_1 $
i.e. Momentum will increase by 100%.
BETA
