A particle of mass 15 kg is moving with a uniform speed 8 ms -1 in x y -plane along the line 3 y=4 x+10, then the magnitude of its angular momentum about the origin in kg - m 2 s -1 is ... ( sin 53°=(4/5))

Question

A particle of mass 15 kg is moving with a uniform speed 8 ms -1 in x y -plane along the line 3 y=4 x+10, then the magnitude of its angular momentum about the origin in kg - m 2 s -1 is ... ( sin 53°=(4/5)) (A) 240 (B) 80 (C) 120 (D) 280

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7516e8bdf2f038e7a08a5d584dd50ae8-.png" /> Angular momentum =m v r1 =15 × 8 ×((10/√32+42)) =15 × 8 × 2=240 kg - m 2 s -1

Q. A particle of mass $15 k g$ is moving with a uniform speed $8 m s^{- 1}$ in $x y$ -plane along the line $3 y = 4 x + 10$ , then the magnitude of its angular momentum about the origin in
$k g - m^{2} s^{- 1}$ is ... $(sin 5 3^{\circ} = \frac{4}{5})$

240

80

120

280

Angular momentum $= m v r_{1}$
$= 15 \times 8 \times (\frac{10}{3 ^{2} + 4 ^{2}})$
$= 15 \times 8 \times 2 = 240 k g - m^{2} s^{- 1}$

Q. A particle of mass 15kg is moving with a uniform speed 8ms−1 in xy -plane along the line 3y=4x+10, then the magnitude of its angular momentum about the origin in kg−m2s−1 is ... (sin53∘=54​)

240

80

120

280

Angular momentum =mvr1​ =15×8×(32+42​10​) =15×8×2=240kg−m2s−1

Q. A particle of mass $15 k g$ is moving with a uniform speed $8 m s^{- 1}$ in $x y$ -plane along the line $3 y = 4 x + 10$ , then the magnitude of its angular momentum about the origin in
$k g - m^{2} s^{- 1}$ is ... $(sin 5 3^{\circ} = \frac{4}{5})$

Angular momentum $= m v r_{1}$
$= 15 \times 8 \times (\frac{10}{3 ^{2} + 4 ^{2}})$
$= 15 \times 8 \times 2 = 240 k g - m^{2} s^{- 1}$