Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Rotational Motion question

2025 · 22 Jan · Shift 1 · Q73
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Rotational Motion
  5. /2025 · 22 Jan · Shift 1 · Q73

Rotational Motion question

2025 · 22 Jan · Shift 1 · Q73

JEE MainPhysicsRotational MotionNumerical+4 / −1
The position vectors of two 1 kg particles, (A) and (B), are given by r→A=(α1t2i^+α2tj^+α3tk^)m and r→B=(β1t^i^+β2t2j^+β3tk^)m, respectively; \overrightarrow{\mathrm{r}}_{\mathrm{A}}=\left(\alpha_1 \mathrm{t}^2 \hat{i}+\alpha_2 \mathrm{t} \hat{j}+\alpha_3 \mathrm{t} \hat{k}\right) \mathrm{m} \text { and } \overrightarrow{\mathrm{r}}_{\mathrm{B}}=\left(\beta_1 \hat{\mathrm{t}} \hat{i}+\beta_2 \mathrm{t}^2 \hat{j}+\beta_3 \mathrm{t} \hat{k}\right) \mathrm{m} \text {, respectively; }rA​=(α1​t2i^+α2​tj^​+α3​tk^)m and rB​=(β1​t^i^+β2​t2j^​+β3​tk^)m, respectively;  (α1=1 m/s2,α2=3nm/s,α3=2 m/s,β1=2 m/s,β2=−1 m/s2,β3=4pm/s)\left(\alpha_1=1 \mathrm{~m} / \mathrm{s}^2, \alpha_2=3 \mathrm{n} \mathrm{m} / \mathrm{s}, \alpha_3=2 \mathrm{~m} / \mathrm{s}, \beta_1=2 \mathrm{~m} / \mathrm{s}, \beta_2=-1 \mathrm{~m} / \mathrm{s}^2, \beta_3=4 \mathrm{pm} / \mathrm{s}\right)(α1​=1 m/s2,α2​=3nm/s,α3​=2 m/s,β1​=2 m/s,β2​=−1 m/s2,β3​=4pm/s), where t is time, n and ppp are constants. At t=1 s,∣VA→∣=∣VB→∣t=1 \mathrm{~s},\left|\overrightarrow{V_A}\right|=\left|\overrightarrow{V_B}\right|t=1 s,​VA​​​=​VB​​​ and velocities V⃗A\vec{V}_AVA​ and V⃗B\vec{V}_BVB​ of the particles are orthogonal to each other. At t=1 st=1 \mathrm{~s}t=1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is Lkgm2 s−1\sqrt{\mathrm{L}} \mathrm{kgm}^2 \mathrm{~s}^{-1}L​kgm2 s−1. The value of L is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 90

  1. Write the position vectors

For particle AAA:

r⃗A=(α1t2i^+α2tj^+α3tk^)\vec r_A=(\alpha_1 t^2\hat i+\alpha_2 t\hat j+\alpha_3 t\hat k)rA​=(α1​t2i^+α2​tj^​+α3​tk^)

Given:

α1=1,α2=3n,α3=2\alpha_1=1,\quad \alpha_2=3n,\quad \alpha_3=2α1​=1,α2​=3n,α3​=2

So,

r⃗A=(t2i^+3ntj^+2tk^)\vec r_A=(t^2\hat i+3nt\hat j+2t\hat k)rA​=(t2i^+3ntj^​+2tk^)

For particle BBB:

r⃗B=(β1ti^+β2t2j^+β3tk^)\vec r_B=(\beta_1 t\hat i+\beta_2 t^2\hat j+\beta_3 t\hat k)rB​=(β1​ti^+β2​t2j^​+β3​tk^)

Given:

β1=2,β2=−1,β3=4p\beta_1=2,\quad \beta_2=-1,\quad \beta_3=4pβ1​=2,β2​=−1,β3​=4p

So,

r⃗B=(2ti^−t2j^+4ptk^)\vec r_B=(2t\hat i-t^2\hat j+4pt\hat k)rB​=(2ti^−t2j^​+4ptk^)
  1. Find velocities

Velocity of AAA:

V⃗A=dr⃗Adt=2ti^+3nj^+2k^\vec V_A=\frac{d\vec r_A}{dt}=2t\hat i+3n\hat j+2\hat kVA​=dtdrA​​=2ti^+3nj^​+2k^

At t=1t=1t=1,

V⃗A=2i^+3nj^+2k^\vec V_A=2\hat i+3n\hat j+2\hat kVA​=2i^+3nj^​+2k^

Velocity of BBB:

V⃗B=dr⃗Bdt=2i^−2tj^+4pk^\vec V_B=\frac{d\vec r_B}{dt}=2\hat i-2t\hat j+4p\hat kVB​=dtdrB​​=2i^−2tj^​+4pk^

At t=1t=1t=1,

V⃗B=2i^−2j^+4pk^\vec V_B=2\hat i-2\hat j+4p\hat kVB​=2i^−2j^​+4pk^
  1. Use orthogonality condition

Given V⃗A\vec V_AVA​ and V⃗B\vec V_BVB​ are perpendicular at t=1t=1t=1:

V⃗A⋅V⃗B=0\vec V_A\cdot \vec V_B=0VA​⋅VB​=0

So,

(2)(2)+(3n)(−2)+(2)(4p)=0(2)(2)+(3n)(-2)+(2)(4p)=0(2)(2)+(3n)(−2)+(2)(4p)=0 4−6n+8p=04-6n+8p=04−6n+8p=0 3n−4p=2...(1)3n-4p=2 \qquad ...(1)3n−4p=2...(1)
  1. Use equal magnitudes condition

Given:

∣V⃗A∣=∣V⃗B∣|\vec V_A|=|\vec V_B|∣VA​∣=∣VB​∣

So,

22+(3n)2+22=22+(−2)2+(4p)22^2+(3n)^2+2^2=2^2+(-2)^2+(4p)^222+(3n)2+22=22+(−2)2+(4p)2 4+9n2+4=4+4+16p24+9n^2+4=4+4+16p^24+9n2+4=4+4+16p2 9n2=16p29n^2=16p^29n2=16p2 3n=±4p3n=\pm 4p3n=±4p

Now combine with equation (1):

3n−4p=23n-4p=23n−4p=2

If 3n=4p3n=4p3n=4p, then

3n−4p=03n-4p=03n−4p=0

which is impossible.

Hence,

3n=−4p3n=-4p3n=−4p

Substitute into (1):

−4p−4p=2-4p-4p=2−4p−4p=2 −8p=2-8p=2−8p=2 p=−14p=-\frac14p=−41​

Then,

3n=1⇒n=133n=1 \Rightarrow n=\frac133n=1⇒n=31​
  1. Find positions at t=1t=1t=1

For AAA:

r⃗A(1)=1i^+3(13)j^+2k^\vec r_A(1)=1\hat i+3\left(\frac13\right)\hat j+2\hat krA​(1)=1i^+3(31​)j^​+2k^ r⃗A(1)=i^+j^+2k^\vec r_A(1)=\hat i+\hat j+2\hat krA​(1)=i^+j^​+2k^

For BBB:

r⃗B(1)=2i^−j^+4(−14)k^\vec r_B(1)=2\hat i-\hat j+4\left(-\frac14\right)\hat krB​(1)=2i^−j^​+4(−41​)k^ r⃗B(1)=2i^−j^−k^\vec r_B(1)=2\hat i-\hat j-\hat krB​(1)=2i^−j^​−k^

Relative position of AAA with respect to BBB:

r⃗AB=r⃗A−r⃗B\vec r_{AB}=\vec r_A-\vec r_BrAB​=rA​−rB​ r⃗AB=(1−2)i^+(1−(−1))j^+(2−(−1))k^\vec r_{AB}=(1-2)\hat i+(1-(-1))\hat j+(2-(-1))\hat krAB​=(1−2)i^+(1−(−1))j^​+(2−(−1))k^ r⃗AB=−i^+2j^+3k^\vec r_{AB}=-\hat i+2\hat j+3\hat krAB​=−i^+2j^​+3k^
  1. Find velocity of A at t=1t=1t=1
V⃗A=2i^+3(13)j^+2k^\vec V_A=2\hat i+3\left(\frac13\right)\hat j+2\hat kVA​=2i^+3(31​)j^​+2k^ V⃗A=2i^+j^+2k^\vec V_A=2\hat i+\hat j+2\hat kVA​=2i^+j^​+2k^

Since mass of particle AAA is 1 kg1\,\text{kg}1kg,

p⃗A=mV⃗A=2i^+j^+2k^\vec p_A=m\vec V_A=2\hat i+\hat j+2\hat kp​A​=mVA​=2i^+j^​+2k^
  1. Angular momentum of A about position of B
L⃗=r⃗AB×p⃗A\vec L=\vec r_{AB}\times \vec p_AL=rAB​×p​A​ L⃗=∣i^j^k^−123212∣\vec L= \begin{vmatrix} \hat i & \hat j & \hat k\\ -1 & 2 & 3\\ 2 & 1 & 2 \end{vmatrix}L=​i^−12​j^​21​k^32​​

Compute:

L⃗=i^(2⋅2−3⋅1)−j^((−1)⋅2−3⋅2)+k^((−1)⋅1−2⋅2)\vec L=\hat i(2\cdot 2-3\cdot 1)-\hat j((-1)\cdot 2-3\cdot 2)+\hat k((-1)\cdot 1-2\cdot 2)L=i^(2⋅2−3⋅1)−j^​((−1)⋅2−3⋅2)+k^((−1)⋅1−2⋅2) L⃗=i^(4−3)−j^(−2−6)+k^(−1−4)\vec L=\hat i(4-3)-\hat j(-2-6)+\hat k(-1-4)L=i^(4−3)−j^​(−2−6)+k^(−1−4) L⃗=i^+8j^−5k^\vec L=\hat i+8\hat j-5\hat kL=i^+8j^​−5k^

Magnitude:

∣L⃗∣=12+82+(−5)2|\vec L|=\sqrt{1^2+8^2+(-5)^2}∣L∣=12+82+(−5)2​ ∣L⃗∣=1+64+25=90|\vec L|=\sqrt{1+64+25}= \sqrt{90}∣L∣=1+64+25​=90​

Given magnitude is L\sqrt{L}L​, so

L=90L=90L=90
  1. Comparison with stored answer

Derived answer: 909090

Stored correct answer: 909090

Hence, they agree.

PreviousNext

More from Rotational Motion

  • The torque due to the force (2i^+j^​+2k^) about the origin, acting on a particle whose position vector is (i^+j^​+k^), would be2025 · MCQ
  • A solid sphere of mass ' m' and radius 'r' is allowed to roll without slipping from the highest point of an inclined plane of length 'L' and makes an angle 30∘ with the horizontal. The speed of the particle at the bottom of…2025 · MCQ
  • A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t)=5t2−8t, where θ(t) is the angular position of the rotating…2025 · MCQ
  • A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination 45∘. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be2025 · MCQ
  • A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be t1​ and t2​, respectively, then2025 · MCQ
  • A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :2025 · MCQ
  • Two iron solid discs of negligible thickness have radii R1​ and R2​ and moment of intertia I1​ and I2​, respectively. For R2​=2R1​, the ratio of I1​ and I2​ would be 1/x, where x=​ .2025 · Numerical
  • The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in…2025 · Numerical