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NEET
Exam
NEET
Subject
Physics
Chapter
Laws of Motion

NEET · Physics · Laws of Motion

On a pulley of mass M hangs a rope with two masses m_ 1 and m_ 2...

On a pulley of mass M hangs a rope with two masses m1\mathrm{m}_{1} and \mathrm{m}_{2}\left(\mathrm{~m}_{1}>\mathrm{m}_{2}\right) tied at the ends as shown in the figure. The pulley rotates without any friction, whereas the friction between the rope and the pulley is large enough to prevent any slipping. Which of the following plots best represents the difference between the tensions in the rope on the two sides of the pulley as a function of the mass of the pulley? On a pulley of mass M hangs a rope with two masses m_ 1 and m_ 2 (~m_ 1 >m_ 2 ) tied at the ends as shown in the

Options

  1. AWhich of the following is correct regarding AIDS causative agent HIV?
  2. BWhich of the following is correct regarding AIDS causative agent HIV?
  3. CWhich of the following is correct regarding AIDS causative agent HIV?
  4. DWhich of the following is correct regarding AIDS causative agent HIV?

Correct answer

C. Which of the following is correct regarding AIDS causative agent HIV?

Step-by-step solution

On a pulley of mass M hangs a rope with two masses m_ 1 and m_ 2 (~m_ 1 >m_ 2 ) tied at the ends as shown in the
mg−T1=m1a......(1)\mathrm{mg}-\mathrm{T}_{1}=\mathrm{m}_{1} \mathrm{a}......\mathrm{(1)} T2−m2 g=m2a......(2)\mathrm{T}_{2}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{2} \mathrm{a}......\mathrm{(2)} (T1−T2)×R=Iα[∵a=rα]=IaR(1)+(2)(m1−m2)g+(T2−T1)=(m1+m2)a(T1−T2)=IR2[(m1−m2)g+(T2−T1)m1+m2]⇒(T1−T2)[1+Ik2( m1+m2)]=Ik2(m1−m2)m1+m2 g⇒T1−T2=[I(m1−m2)gI+(m1+m2)R2]=12MR2( m1−m2)g(12MR2+(m1+m2)R2)T1−T2=M(m1−m2)gM+2( m1+m2)\begin{array}{l} \left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) \times \mathrm{R}=\mathrm{I} \alpha \quad[\because \mathrm{a}=\mathrm{r} \alpha] \\ =\frac{\mathrm{Ia}}{\mathrm{R}} \\ (1)+(2) \\ \left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}+\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{a} \\ \left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\frac{\mathrm{I}}{\mathrm{R}^{2}}\left[\frac{\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}+\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right] \\ \Rightarrow\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)\left[1+\frac{\mathrm{I}}{\mathrm{k}^{2}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}\right] \\ =\frac{\mathrm{I}}{\mathrm{k}^{2}} \frac{\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)}{\mathrm{m}_{1}+\mathrm{m}_{2}} \mathrm{~g} \\ \Rightarrow \mathrm{T}_{1}-\mathrm{T}_{2}=\left[\frac{\mathrm{I}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}}{\mathrm{I}+\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{R}^{2}}\right] \\ =\frac{\frac{1}{2} \mathrm{MR}^{2}\left(\mathrm{~m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}}{\left(\frac{1}{2} \mathrm{MR}^{2}+\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{R}^{2}\right)} \\ \mathrm{T}_{1}-\mathrm{T}_{2}=\frac{\mathrm{M}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}}{\mathrm{M}+2\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)} \end{array}
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