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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 subscript blank superscript 3 end superscript comma M a s s of cart equals M equals 10 text end text k g. Initially left parenthesis t equals 0 right parenthesis the force on the cart is equal to :

  1. 20 N    
  2. 40 N    
  3. 80 N    
  4. zero    

The correct answer is: 40 N


    table row cell F equals rho A open parentheses V subscript 0 end subscript minus 0 close parentheses to the power of 2 end exponent open square brackets 1 minus cos invisible function application 180 to the power of ring operator end exponent close square brackets end cell row cell equals 2 p A v to the power of 2 end exponent equals 2 cross times 1000 cross times 2 cross times 10 to the power of negative 4 end exponent cross times 10 cross times 10 equals 40 N end cell end table

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