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AIIMS2017Physics-Work, Energy and Power

AIIMS 2017 Physics Work Done by a Variable Force MCQ Question

Type: MCQ-numerical-Medium-Class 11

A conductor lies along the x-axis at −1.5≤Z≤1.5 m-1.5 \le Z \le 1.5\text{ m} carries a fixed current of 10.0 A10.0\text{ A} in −az-a_z direction as shown in the figure for the field B=3×10−4e−0.2xay TB = 3 \times 10^{-4} e^{-0.2x} a_y\text{ T}, the total power required to move the conductor at constant speed to x=2.0 m,y=0 mx = 2.0\text{ m}, y = 0\text{ m} in 5×10−3 s5 \times 10^{-3}\text{ s} is (Assume parallel motion along the x-axis)

Question diagram
A

1.57 W

B

2.97 W

C

4.45 W

D

9.87 W

Correct Answer

Option B

Detailed Explanation

Option B, W=∫02BII dxW = \int_{0}^{2} B_{II} \, dx, is correct as it represents the work done in a system where BIIB_{II} signifies a specific force or energy density relevant to the context of the problem. The other options are incorrect because option A lacks the specific context of BIIB_{II}, while options C and D introduce unnecessary constants and exponential functions that do not align with the standard definition of work done in a physical system. Understanding the integral formulation for work, W=∫F dxW = \int F \, dx, is crucial for applying it correctly in various scenarios, including those involving variable forces or energy densities.

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