NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
269401
Mass of thin long metal rod is\(2 \mathrm{~kg}\) and its moment of inertia about an axis perpendicular to the length of rod and passing through its one end is \(0.5 \mathrm{~kg} \mathrm{~m}^{2}\). Its radius of gyration is
269414
An automobile engine develops\(100 \mathrm{KW}\) when rotating at a speed of \(1800 \mathrm{rev} / \mathrm{min}\). The torque it delivers ( in \(\mathrm{N}-\mathrm{m}\) )
1 350
2 440
3 531
4 628
Explanation:
\(p=\pi \omega \quad\)
Rotational Motion
269415
An electric motor exerts a constant torque\(5 \mathrm{Nm}\) on a fly wheel by which it is rotated at the rate of \(420 \mathrm{rpm}\) The power of motor is
1 110watt
2 150watt
3 220watt
4 300watt
Explanation:
\(p=\pi \omega\)
Rotational Motion
269445
The ratio of moments of inertia of a solid sphere about axes passing through itscentre and tangent respectively is
1 \(2: 5\)
2 \(2: 7\)
3 \(5: 2\)
4 \(7: 2\)
Explanation:
\(\frac{I_{\text {centre }}}{I_{\text {tanget }}}=\frac{\frac{2}{5} M R^{2}}{\frac{7}{5} M R^{2}}=\frac{2}{7}\)
269401
Mass of thin long metal rod is\(2 \mathrm{~kg}\) and its moment of inertia about an axis perpendicular to the length of rod and passing through its one end is \(0.5 \mathrm{~kg} \mathrm{~m}^{2}\). Its radius of gyration is
269414
An automobile engine develops\(100 \mathrm{KW}\) when rotating at a speed of \(1800 \mathrm{rev} / \mathrm{min}\). The torque it delivers ( in \(\mathrm{N}-\mathrm{m}\) )
1 350
2 440
3 531
4 628
Explanation:
\(p=\pi \omega \quad\)
Rotational Motion
269415
An electric motor exerts a constant torque\(5 \mathrm{Nm}\) on a fly wheel by which it is rotated at the rate of \(420 \mathrm{rpm}\) The power of motor is
1 110watt
2 150watt
3 220watt
4 300watt
Explanation:
\(p=\pi \omega\)
Rotational Motion
269445
The ratio of moments of inertia of a solid sphere about axes passing through itscentre and tangent respectively is
1 \(2: 5\)
2 \(2: 7\)
3 \(5: 2\)
4 \(7: 2\)
Explanation:
\(\frac{I_{\text {centre }}}{I_{\text {tanget }}}=\frac{\frac{2}{5} M R^{2}}{\frac{7}{5} M R^{2}}=\frac{2}{7}\)
269401
Mass of thin long metal rod is\(2 \mathrm{~kg}\) and its moment of inertia about an axis perpendicular to the length of rod and passing through its one end is \(0.5 \mathrm{~kg} \mathrm{~m}^{2}\). Its radius of gyration is
269414
An automobile engine develops\(100 \mathrm{KW}\) when rotating at a speed of \(1800 \mathrm{rev} / \mathrm{min}\). The torque it delivers ( in \(\mathrm{N}-\mathrm{m}\) )
1 350
2 440
3 531
4 628
Explanation:
\(p=\pi \omega \quad\)
Rotational Motion
269415
An electric motor exerts a constant torque\(5 \mathrm{Nm}\) on a fly wheel by which it is rotated at the rate of \(420 \mathrm{rpm}\) The power of motor is
1 110watt
2 150watt
3 220watt
4 300watt
Explanation:
\(p=\pi \omega\)
Rotational Motion
269445
The ratio of moments of inertia of a solid sphere about axes passing through itscentre and tangent respectively is
1 \(2: 5\)
2 \(2: 7\)
3 \(5: 2\)
4 \(7: 2\)
Explanation:
\(\frac{I_{\text {centre }}}{I_{\text {tanget }}}=\frac{\frac{2}{5} M R^{2}}{\frac{7}{5} M R^{2}}=\frac{2}{7}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Rotational Motion
269401
Mass of thin long metal rod is\(2 \mathrm{~kg}\) and its moment of inertia about an axis perpendicular to the length of rod and passing through its one end is \(0.5 \mathrm{~kg} \mathrm{~m}^{2}\). Its radius of gyration is
269414
An automobile engine develops\(100 \mathrm{KW}\) when rotating at a speed of \(1800 \mathrm{rev} / \mathrm{min}\). The torque it delivers ( in \(\mathrm{N}-\mathrm{m}\) )
1 350
2 440
3 531
4 628
Explanation:
\(p=\pi \omega \quad\)
Rotational Motion
269415
An electric motor exerts a constant torque\(5 \mathrm{Nm}\) on a fly wheel by which it is rotated at the rate of \(420 \mathrm{rpm}\) The power of motor is
1 110watt
2 150watt
3 220watt
4 300watt
Explanation:
\(p=\pi \omega\)
Rotational Motion
269445
The ratio of moments of inertia of a solid sphere about axes passing through itscentre and tangent respectively is
1 \(2: 5\)
2 \(2: 7\)
3 \(5: 2\)
4 \(7: 2\)
Explanation:
\(\frac{I_{\text {centre }}}{I_{\text {tanget }}}=\frac{\frac{2}{5} M R^{2}}{\frac{7}{5} M R^{2}}=\frac{2}{7}\)