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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Physics
Chapter
Gravitation

JEE Advanced 2010 · Paper 1 · Physics · Gravitation

A thin uniform annular disc (see figure) of mass M has outer radius...

A thin uniform annular disc (see figure) of mass MM has outer radius 4R4 R and inner radius 3R3 R. The work required to take a unit mass from point PP on its axis to infinity is A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3 R . The work required to

Options

  1. A2GM7R(42−5)\frac{2 G M}{7 R}(4 \sqrt{2}-5)
  2. B−2GM7R(42−5)-\frac{2 G M}{7 R}(4 \sqrt{2}-5)
  3. CGM4R\frac{G M}{4 R}
  4. D2GM5R(2−1)\frac{2 G M}{5 R}(\sqrt{2}-1)

Correct answer

A. 2GM7R(42−5)\frac{2 G M}{7 R}(4 \sqrt{2}-5)

Step-by-step solution

W=\DeltaU=Uf−Ui=U∞−Up=−UP=−mVP=−VP( as m=1)\begin{aligned}W=\DeltaU & =U_f-U_i=U_{\infty}-U_p \\ & =-U_P \\ & =-m V_P \\ & =-V_P \quad(\text{ as }m=1) \end{aligned}Potential at point PP will be obtained by integration as given below <imgsrc="/api/proxy−image?url=https<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_advanced%252FjJmBXKLu523-AnbaAkWokVGroIl4PK3pdsvhCw-qfao.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2FjJmBXKLu523-AnbaAkWokVGroIl4PK3pdsvhCw-qfao.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2FjJmBXKLu523-AnbaAkWokVGroIl4PK3pdsvhCw-qfao.original.fullsize.png?alt=media"> \\ Let dMd M be the mass of small ring as shown dM=Mπ(4R)2−π(3R)2(2\pir)dr=2Mrdr7R2dVP=−G⋅dM16R2+r2=−2GM7R2∫3R4Rr16R2+r2\cdotdr=−2GM7R(42−5)∴W=+2GM7R(42−5)\begin{aligned}d M & =\frac{M}{\pi(4 R)^2-\pi(3 R)^2}(2 \pir) d r \\ & =\frac{2 M r d r}{7 R^2}\\ d V_P & =-\frac{G \cdot d M}{\sqrt{16 R^2+r^2}}\\ & =-\frac{2 G M}{7 R^2} \int_{3 R}^{4 R} \frac{r}{\sqrt{16 R^2+r^2}} \cdotd r \\ & =-\frac{2 G M}{7 R}(4 \sqrt{2}-5) \\ \therefore \quad W & =+\frac{2 G M}{7 R}(4 \sqrt{2}-5) \end{aligned}\therefore$ correct option is (a).
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