A particle acted on by constant forces 4^i+^j−3^k and 3^i+^j−^k is displaced from the point ^i+2^j+3^k to the point 5^i+4^j+^k. The total work done by the forces is
A
20 units
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B
30 units
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C
40 units
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D
50 units
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Solution
The correct option is C 40 units Total work done = [(4^i+^j−3^k)+(3^i+^j−^k)].[(5^i+4^j+^k)−(^i+2^j+3^k)] =(7^i+2^j−4^k).(4^i+2^j−2^k)=28+4+8=40