144059
The acceleration of an object moving in a circle of radius \(R\) with uniform speed \(v\) is
1 \(\frac{v^{2}}{R}\)
2 \(\frac{v^{2}}{2 R}\)
3 \(\frac{2 v^{2}}{R}\)
4 \(\frac{3 v^{2}}{2 R}\)
Explanation:
A We know that, \(\mathrm{v}=\omega \mathrm{r}\) \(\mathrm{a}_{\mathrm{C}}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\omega \mathrm{r}) \quad\left(\mathrm{a}_{\mathrm{C}}=\right.\) centripetal acceleration \()\) \(\mathrm{a}_{\mathrm{C}}=\omega \frac{\mathrm{dr}}{\mathrm{dt}} \quad(\omega\) is constant \()\) \(\mathrm{a}_{\mathrm{C}} =\omega \times \mathrm{v} \quad\left\{\because \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{v}\right\}\) \(=\frac{\mathrm{v}}{\mathrm{R}} \cdot \mathrm{v}\) \(\mathrm{a}_{\mathrm{C}} =\frac{\mathrm{v}^{2}}{\mathrm{R}}\)
J and K-CET-2011
Motion in Plane
144060
If a car is to travel with a speed \(v\) along the frictionless, banked circular track of radius \(r\), the required angle of banking so that the car does skid is
A Let \(\theta\) is the angle of banking In horizontal, \(\mathrm{mg} \sin \theta=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) In vertical, \(\mathrm{mg} \cos \theta=\mathrm{mg}\) Dividing equation (i) by equation (ii) \(\frac{\mathrm{mg} \sin \theta}{\mathrm{mg} \cos \theta}=\frac{\mathrm{mv}^{2}}{\mathrm{r} \times \mathrm{mg}}\) \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \theta=\tan ^{-1}\left(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\right)\)
J and K-CET-2010
Motion in Plane
144063
Angle of banking for a vehicle speed of \(10 \mathrm{~m} / \mathrm{s}\) for a radius of curvature \(10 \mathrm{~m}\) is (assume \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )
1 \(30^{\circ}\)
2 \(\tan ^{-1}(1 / 2)\)
3 \(60^{\circ}\)
4 \(45^{\circ}\)
Explanation:
D Given, \(\mathrm{v}=10 \mathrm{~m} / \mathrm{s}, \mathrm{r}=10 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know, Angle of banking, \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \tan \theta=\frac{10 \times 10}{10 \times 10}=1\) \(\tan \theta=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
J and K-CET-1998
Motion in Plane
144064
A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of \(0.5 \mathrm{~m} / \mathrm{s}\). What is the height of the plane of circle from vertex of the funnel?
1 \(0.25 \mathrm{~cm}\)
2 \(2 \mathrm{~cm}\)
3 \(4 \mathrm{~cm}\)
4 \(2.5 \mathrm{~cm}\)
Explanation:
D Given that, particle is moving in circular path So, figure \(\mathrm{mg}=\mathrm{R} \sin \theta\) \(\frac{m v^{2}}{r}=R \cos \theta\) From equation (i) and (ii) \(\therefore \quad \tan \theta=\frac{\mathrm{rg}}{\mathrm{v}^{2}} \text { but } \tan \theta=\frac{\mathrm{r}}{\mathrm{h}}\) \(\mathrm{h}=\frac{\mathrm{v}^{2}}{\mathrm{~g}}=\frac{(0.5)^{2}}{10}=0.025 \mathrm{~m}=2.5 \mathrm{~cm}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in Plane
144059
The acceleration of an object moving in a circle of radius \(R\) with uniform speed \(v\) is
1 \(\frac{v^{2}}{R}\)
2 \(\frac{v^{2}}{2 R}\)
3 \(\frac{2 v^{2}}{R}\)
4 \(\frac{3 v^{2}}{2 R}\)
Explanation:
A We know that, \(\mathrm{v}=\omega \mathrm{r}\) \(\mathrm{a}_{\mathrm{C}}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\omega \mathrm{r}) \quad\left(\mathrm{a}_{\mathrm{C}}=\right.\) centripetal acceleration \()\) \(\mathrm{a}_{\mathrm{C}}=\omega \frac{\mathrm{dr}}{\mathrm{dt}} \quad(\omega\) is constant \()\) \(\mathrm{a}_{\mathrm{C}} =\omega \times \mathrm{v} \quad\left\{\because \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{v}\right\}\) \(=\frac{\mathrm{v}}{\mathrm{R}} \cdot \mathrm{v}\) \(\mathrm{a}_{\mathrm{C}} =\frac{\mathrm{v}^{2}}{\mathrm{R}}\)
J and K-CET-2011
Motion in Plane
144060
If a car is to travel with a speed \(v\) along the frictionless, banked circular track of radius \(r\), the required angle of banking so that the car does skid is
A Let \(\theta\) is the angle of banking In horizontal, \(\mathrm{mg} \sin \theta=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) In vertical, \(\mathrm{mg} \cos \theta=\mathrm{mg}\) Dividing equation (i) by equation (ii) \(\frac{\mathrm{mg} \sin \theta}{\mathrm{mg} \cos \theta}=\frac{\mathrm{mv}^{2}}{\mathrm{r} \times \mathrm{mg}}\) \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \theta=\tan ^{-1}\left(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\right)\)
J and K-CET-2010
Motion in Plane
144063
Angle of banking for a vehicle speed of \(10 \mathrm{~m} / \mathrm{s}\) for a radius of curvature \(10 \mathrm{~m}\) is (assume \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )
1 \(30^{\circ}\)
2 \(\tan ^{-1}(1 / 2)\)
3 \(60^{\circ}\)
4 \(45^{\circ}\)
Explanation:
D Given, \(\mathrm{v}=10 \mathrm{~m} / \mathrm{s}, \mathrm{r}=10 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know, Angle of banking, \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \tan \theta=\frac{10 \times 10}{10 \times 10}=1\) \(\tan \theta=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
J and K-CET-1998
Motion in Plane
144064
A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of \(0.5 \mathrm{~m} / \mathrm{s}\). What is the height of the plane of circle from vertex of the funnel?
1 \(0.25 \mathrm{~cm}\)
2 \(2 \mathrm{~cm}\)
3 \(4 \mathrm{~cm}\)
4 \(2.5 \mathrm{~cm}\)
Explanation:
D Given that, particle is moving in circular path So, figure \(\mathrm{mg}=\mathrm{R} \sin \theta\) \(\frac{m v^{2}}{r}=R \cos \theta\) From equation (i) and (ii) \(\therefore \quad \tan \theta=\frac{\mathrm{rg}}{\mathrm{v}^{2}} \text { but } \tan \theta=\frac{\mathrm{r}}{\mathrm{h}}\) \(\mathrm{h}=\frac{\mathrm{v}^{2}}{\mathrm{~g}}=\frac{(0.5)^{2}}{10}=0.025 \mathrm{~m}=2.5 \mathrm{~cm}\)
144059
The acceleration of an object moving in a circle of radius \(R\) with uniform speed \(v\) is
1 \(\frac{v^{2}}{R}\)
2 \(\frac{v^{2}}{2 R}\)
3 \(\frac{2 v^{2}}{R}\)
4 \(\frac{3 v^{2}}{2 R}\)
Explanation:
A We know that, \(\mathrm{v}=\omega \mathrm{r}\) \(\mathrm{a}_{\mathrm{C}}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\omega \mathrm{r}) \quad\left(\mathrm{a}_{\mathrm{C}}=\right.\) centripetal acceleration \()\) \(\mathrm{a}_{\mathrm{C}}=\omega \frac{\mathrm{dr}}{\mathrm{dt}} \quad(\omega\) is constant \()\) \(\mathrm{a}_{\mathrm{C}} =\omega \times \mathrm{v} \quad\left\{\because \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{v}\right\}\) \(=\frac{\mathrm{v}}{\mathrm{R}} \cdot \mathrm{v}\) \(\mathrm{a}_{\mathrm{C}} =\frac{\mathrm{v}^{2}}{\mathrm{R}}\)
J and K-CET-2011
Motion in Plane
144060
If a car is to travel with a speed \(v\) along the frictionless, banked circular track of radius \(r\), the required angle of banking so that the car does skid is
A Let \(\theta\) is the angle of banking In horizontal, \(\mathrm{mg} \sin \theta=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) In vertical, \(\mathrm{mg} \cos \theta=\mathrm{mg}\) Dividing equation (i) by equation (ii) \(\frac{\mathrm{mg} \sin \theta}{\mathrm{mg} \cos \theta}=\frac{\mathrm{mv}^{2}}{\mathrm{r} \times \mathrm{mg}}\) \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \theta=\tan ^{-1}\left(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\right)\)
J and K-CET-2010
Motion in Plane
144063
Angle of banking for a vehicle speed of \(10 \mathrm{~m} / \mathrm{s}\) for a radius of curvature \(10 \mathrm{~m}\) is (assume \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )
1 \(30^{\circ}\)
2 \(\tan ^{-1}(1 / 2)\)
3 \(60^{\circ}\)
4 \(45^{\circ}\)
Explanation:
D Given, \(\mathrm{v}=10 \mathrm{~m} / \mathrm{s}, \mathrm{r}=10 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know, Angle of banking, \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \tan \theta=\frac{10 \times 10}{10 \times 10}=1\) \(\tan \theta=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
J and K-CET-1998
Motion in Plane
144064
A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of \(0.5 \mathrm{~m} / \mathrm{s}\). What is the height of the plane of circle from vertex of the funnel?
1 \(0.25 \mathrm{~cm}\)
2 \(2 \mathrm{~cm}\)
3 \(4 \mathrm{~cm}\)
4 \(2.5 \mathrm{~cm}\)
Explanation:
D Given that, particle is moving in circular path So, figure \(\mathrm{mg}=\mathrm{R} \sin \theta\) \(\frac{m v^{2}}{r}=R \cos \theta\) From equation (i) and (ii) \(\therefore \quad \tan \theta=\frac{\mathrm{rg}}{\mathrm{v}^{2}} \text { but } \tan \theta=\frac{\mathrm{r}}{\mathrm{h}}\) \(\mathrm{h}=\frac{\mathrm{v}^{2}}{\mathrm{~g}}=\frac{(0.5)^{2}}{10}=0.025 \mathrm{~m}=2.5 \mathrm{~cm}\)
144059
The acceleration of an object moving in a circle of radius \(R\) with uniform speed \(v\) is
1 \(\frac{v^{2}}{R}\)
2 \(\frac{v^{2}}{2 R}\)
3 \(\frac{2 v^{2}}{R}\)
4 \(\frac{3 v^{2}}{2 R}\)
Explanation:
A We know that, \(\mathrm{v}=\omega \mathrm{r}\) \(\mathrm{a}_{\mathrm{C}}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\omega \mathrm{r}) \quad\left(\mathrm{a}_{\mathrm{C}}=\right.\) centripetal acceleration \()\) \(\mathrm{a}_{\mathrm{C}}=\omega \frac{\mathrm{dr}}{\mathrm{dt}} \quad(\omega\) is constant \()\) \(\mathrm{a}_{\mathrm{C}} =\omega \times \mathrm{v} \quad\left\{\because \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{v}\right\}\) \(=\frac{\mathrm{v}}{\mathrm{R}} \cdot \mathrm{v}\) \(\mathrm{a}_{\mathrm{C}} =\frac{\mathrm{v}^{2}}{\mathrm{R}}\)
J and K-CET-2011
Motion in Plane
144060
If a car is to travel with a speed \(v\) along the frictionless, banked circular track of radius \(r\), the required angle of banking so that the car does skid is
A Let \(\theta\) is the angle of banking In horizontal, \(\mathrm{mg} \sin \theta=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) In vertical, \(\mathrm{mg} \cos \theta=\mathrm{mg}\) Dividing equation (i) by equation (ii) \(\frac{\mathrm{mg} \sin \theta}{\mathrm{mg} \cos \theta}=\frac{\mathrm{mv}^{2}}{\mathrm{r} \times \mathrm{mg}}\) \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \theta=\tan ^{-1}\left(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\right)\)
J and K-CET-2010
Motion in Plane
144063
Angle of banking for a vehicle speed of \(10 \mathrm{~m} / \mathrm{s}\) for a radius of curvature \(10 \mathrm{~m}\) is (assume \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )
1 \(30^{\circ}\)
2 \(\tan ^{-1}(1 / 2)\)
3 \(60^{\circ}\)
4 \(45^{\circ}\)
Explanation:
D Given, \(\mathrm{v}=10 \mathrm{~m} / \mathrm{s}, \mathrm{r}=10 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know, Angle of banking, \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \tan \theta=\frac{10 \times 10}{10 \times 10}=1\) \(\tan \theta=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
J and K-CET-1998
Motion in Plane
144064
A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of \(0.5 \mathrm{~m} / \mathrm{s}\). What is the height of the plane of circle from vertex of the funnel?
1 \(0.25 \mathrm{~cm}\)
2 \(2 \mathrm{~cm}\)
3 \(4 \mathrm{~cm}\)
4 \(2.5 \mathrm{~cm}\)
Explanation:
D Given that, particle is moving in circular path So, figure \(\mathrm{mg}=\mathrm{R} \sin \theta\) \(\frac{m v^{2}}{r}=R \cos \theta\) From equation (i) and (ii) \(\therefore \quad \tan \theta=\frac{\mathrm{rg}}{\mathrm{v}^{2}} \text { but } \tan \theta=\frac{\mathrm{r}}{\mathrm{h}}\) \(\mathrm{h}=\frac{\mathrm{v}^{2}}{\mathrm{~g}}=\frac{(0.5)^{2}}{10}=0.025 \mathrm{~m}=2.5 \mathrm{~cm}\)
