Q.30. A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire c

Q: What is the correct answer for JEE Main Physics Q30?

The correct answer is A. Sol. Mass of lift $m = 1600 \text{ kg}$. Maximum stress $\sigma_{\text{max}} = 4 \times 10^8 \text{ N/m}^2$. Radius of wire $r = 4 \text{ mm} = 4 \times 10^{-3} \text{ m}$. Area of cross-section of wire $A = \pi r^2 = 3.14 \times (4 \times 10^{-3})^2 = 3.14 \times 16 \times 10^{-6} = 50.24 \times 10^{-6} \text{ m}^2$. Maximum tension the wire can withstand is $T_{\text{max}} = \sigma_{\text{max}}

Q: Q.30. A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is and its radius is 4 mm, then maximum acceleration the lift can take is ________ .…

The correct answer is A. Sol. Mass of lift $m = 1600 \text{ kg}$. Maximum stress $\sigma_{\text{max}} = 4 \times 10^8 \text{ N/m}^2$. Radius of wire $r = 4 \text{ mm} = 4 \times 10^{-3} \text{ m}$. Area of cross-section of wire $A = \pi r^2 = 3.14 \times (4 \times 10^{-3})^2 = 3.14 \times 16 \times 10^{-6} = 50.24 \times 10^{-6} \text{ m}^2$. Maximum tension the wire can withstand is $T_{\text{max}} = \sigma_{\text{max}}

Q: What are the options for this JEE Main Physics question?

A. 56 · B. 89 · C. 32 · D. 16

Question

Q.30. A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is  and its radius is 4 mm, then maximum acceleration the lift can take is ________ . (take  and )

FirstInTest · Accepted Answer

Sol. Mass of lift $m = 1600 	ext{ kg}$. Maximum stress $\sigma_{	ext{max}} = 4 	imes 10^8 	ext{ N/m}^2$. Radius of wire $r = 4 	ext{ mm} = 4 	imes 10^{-3} 	ext{ m}$. Area of cross-section of wire $A = \pi r^2 = 3.14 	imes (4 	imes 10^{-3})^2 = 3.14 	imes 16 	imes 10^{-6} = 50.24 	imes 10^{-6} 	ext{ m}^2$. Maximum tension the wire can withstand is $T_{	ext{max}} = \sigma_{	ext{max}} 	imes A = (4 	imes 10^8 	ext{ N/m}^2) 	imes (50.24 	imes 10^{-6} 	ext{ m}^2) = 200.96 	imes 10^2 = 20096 	ext{ N}$. When the lift accelerates upwards with acceleration $a$, the equation of motion is $T - mg = ma$. So, $T = m(g+a)$. For maximum acceleration, $T$ must be $T_{	ext{max}}$. $T_{	ext{max}} = m(g+a_{	ext{max}})$. $20096 = 1600 (10 + a_{	ext{max}})$. $10 + a_{	ext{max}} = \frac{20096}{1600} = 12.56$. $a_{	ext{max}} = 12.56 - 10 = 2.56 	ext{ m/s}^2$.

Answer

56

Answer

89

Answer

32

Answer

16

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