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on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses V subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :

Given cross-section area of jet equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent velocity of jet V subscript 0 end subscript equals 10 text end text m divided by s e c, density of liquid equals 1000 text end text k g divided by m to the power of 3 end exponent comma M a s s of cart equals M equals 10 text end text k g. The power supplied to the cart, when its velocity becomes 5 m/sec., is equal to :

  1. 100 W    
  2. 25 W    
  3. 50 W    
  4. 200 W    

The correct answer is: 50 W


    table row cell F equals 2 rho A open parentheses V subscript 0 end subscript minus u close parentheses to the power of 2 end exponent equals 2 cross times 10 to the power of 3 end exponent cross times 2 cross times 10 to the power of negative 4 end exponent cross times 25 equals 10 N end cell row cell P equals F times u equals 10 cross times 5 equals 50 W end cell end table

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