The magnitude of force acting on a body varies with time as shown in the figure. The magnitude of total impulse of the force on the body from t= 0 to t=16μs is-

Question

The magnitude of force acting on a body varies with time as shown in the figure. The magnitude of total impulse of the force on the body from t= 0 to t=16μs is- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-8aobrnfbnxyucd7x.jpg" /> (A) 6× 10- 3Ns (B) 5.8× 10- 3Ns (C) 8.4× 10- 3Ns (D) 7.6× 10- 3Ns

Tardigrade Admin · Accepted Answer

Impulse = displaystyle ∫ F d t = area under curve Area =(200 × 4 + (1/2) × 2 × (200 + 800) + (1/2) × 10 × 800)× 10- 6 Area =(800 + 1000 + 4000)× 10- 6 Area =5800× 10- 6=5.8× 10- 3 textNs

Q. The magnitude of force acting on a body varies with time as shown in the figure. The magnitude of total impulse of the force on the body from $t = 0$ to $t = 16 μ s$ is-

$6 \times 1 0^{- 3} N s$

$5.8 \times 1 0^{- 3} N s$

$8.4 \times 1 0^{- 3} N s$

$7.6 \times 1 0^{- 3} N s$

Impulse $= \int F d t$ = area under curve
Area $= (200 \times 4 + \frac{1}{2} \times 2 \times (200 + 800) + \frac{1}{2} \times 10 \times 800) \times 1 0^{- 6}$
Area $= (800 + 1000 + 4000) \times 1 0^{- 6}$
Area $= 5800 \times 1 0^{- 6} = 5.8 \times 1 0^{- 3} Ns$

Q. The magnitude of force acting on a body varies with time as shown in the figure. The magnitude of total impulse of the force on the body from t=0 to t=16μs is-

6×10−3Ns

5.8×10−3Ns

8.4×10−3Ns

7.6×10−3Ns