NEET Test Series from KOTA - 10 Papers In MS WORD
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LAWS OF MOTION (ADDITIONAL)
371872
The maximum and minimum tensions in a string whirling in a circle of radius \(2.5 \mathrm{~m}\) are in the ratio \(5: 3\). Then, its velocity is m.s :
1 7
2 \(\sqrt{98}\)
3 \(\sqrt{4.9}\)
4 \(\sqrt{490}\)
Explanation:
B Given, radius of circle (r) \(=2.5 \mathrm{~m}\) It is assumed that particle although moving in a vertical loop but its speed is constant. Tension of lowest point- \(\mathrm{T}_{\max }=\frac{\mathrm{mv}^{2}}{\mathrm{r}}+\mathrm{mg}\) Tension of highest point- \(\mathrm{T}_{\text {min. }}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}\) Dividing equation (i) by (ii), we get- \(\frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{\frac{\mathrm{mv}}{\mathrm{r}}+\mathrm{mg}}{\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}}=\frac{5}{3} \quad\left[\frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{5}{3} \text { (given) }\right]\) \(\frac{\mathrm{v}^{2}+\mathrm{gr}}{\mathrm{v}^{2}-\mathrm{gr}}=\frac{5}{3}\) \(\mathrm{v}=\sqrt{4 \mathrm{gr}}\) \(\mathrm{v}=\sqrt{4 \times 9.8 \times 2.5}\) \(\mathrm{v}=\sqrt{98} \mathrm{~ms}^{-1}\)
AP EAMCET-25.09.2020
LAWS OF MOTION (ADDITIONAL)
371873
Two bodies of mass \(4 \mathrm{~kg}\) and \(6 \mathrm{~kg}\) are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity \(g\) is
1 \(g / 2\)
2 \(g / 5\)
3 \(g / 10\)
4 \(\mathrm{g}\)
Explanation:
B Given, \(\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=6 \mathrm{~kg}\) For block A - The force balancing equation for \(4 \mathrm{~kg}\) block, \(\mathrm{T}-\mathrm{m}_{1} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}\) \(\mathrm{T}-4 \mathrm{~g}=4 \mathrm{a}\) For block B - Force balancing equation for \(6 \mathrm{~kg}\) block, \(\mathrm{m}_{2} \mathrm{~g}-\mathrm{T}=\mathrm{m}_{2} \mathrm{a}\) \(6 \mathrm{~g}-\mathrm{T}=6 \mathrm{a}\) Adding equation (i) and (ii), we get - \(2 g=10 a\) \(a=g / 5\)
NEET- (Sep.) 2020
LAWS OF MOTION (ADDITIONAL)
371874
A body of mass \(2 \mathrm{~kg}\) is acted upon by two each of magnitude \(1 \mathrm{~N}\) and inclined at \(60^{\circ}\) with each other. The acceleration of the body in \(\mathrm{m} / \mathrm{s}^{2}\) is \(\left(\cos 60^{\circ}=0.5\right)\)
371875
A see-saw of length \(6 \mathrm{~m}\) is pivoted at its centre. A child of mass \(20 \mathrm{~kg}\) is sitting at one of its ends. Where should another child, of mass 30 \(\mathrm{kg}\), sit on the other end from the centre of seesaw, so that it is balanced ?
1 \(1 \mathrm{~m}\)
2 \(3 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
C Let, 'n' be the distance of B from the fulcrum. \(\mathrm{F}_{1}=20 \mathrm{~g}\) and \(\mathrm{F}_{2}=30 \mathrm{~g}\) For equilibrium, \(20 \times 3=30 \times x\) \(\mathrm{x}=\frac{60}{30}\) \(\mathrm{x}=2 \mathrm{~m}\)
371872
The maximum and minimum tensions in a string whirling in a circle of radius \(2.5 \mathrm{~m}\) are in the ratio \(5: 3\). Then, its velocity is m.s :
1 7
2 \(\sqrt{98}\)
3 \(\sqrt{4.9}\)
4 \(\sqrt{490}\)
Explanation:
B Given, radius of circle (r) \(=2.5 \mathrm{~m}\) It is assumed that particle although moving in a vertical loop but its speed is constant. Tension of lowest point- \(\mathrm{T}_{\max }=\frac{\mathrm{mv}^{2}}{\mathrm{r}}+\mathrm{mg}\) Tension of highest point- \(\mathrm{T}_{\text {min. }}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}\) Dividing equation (i) by (ii), we get- \(\frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{\frac{\mathrm{mv}}{\mathrm{r}}+\mathrm{mg}}{\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}}=\frac{5}{3} \quad\left[\frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{5}{3} \text { (given) }\right]\) \(\frac{\mathrm{v}^{2}+\mathrm{gr}}{\mathrm{v}^{2}-\mathrm{gr}}=\frac{5}{3}\) \(\mathrm{v}=\sqrt{4 \mathrm{gr}}\) \(\mathrm{v}=\sqrt{4 \times 9.8 \times 2.5}\) \(\mathrm{v}=\sqrt{98} \mathrm{~ms}^{-1}\)
AP EAMCET-25.09.2020
LAWS OF MOTION (ADDITIONAL)
371873
Two bodies of mass \(4 \mathrm{~kg}\) and \(6 \mathrm{~kg}\) are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity \(g\) is
1 \(g / 2\)
2 \(g / 5\)
3 \(g / 10\)
4 \(\mathrm{g}\)
Explanation:
B Given, \(\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=6 \mathrm{~kg}\) For block A - The force balancing equation for \(4 \mathrm{~kg}\) block, \(\mathrm{T}-\mathrm{m}_{1} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}\) \(\mathrm{T}-4 \mathrm{~g}=4 \mathrm{a}\) For block B - Force balancing equation for \(6 \mathrm{~kg}\) block, \(\mathrm{m}_{2} \mathrm{~g}-\mathrm{T}=\mathrm{m}_{2} \mathrm{a}\) \(6 \mathrm{~g}-\mathrm{T}=6 \mathrm{a}\) Adding equation (i) and (ii), we get - \(2 g=10 a\) \(a=g / 5\)
NEET- (Sep.) 2020
LAWS OF MOTION (ADDITIONAL)
371874
A body of mass \(2 \mathrm{~kg}\) is acted upon by two each of magnitude \(1 \mathrm{~N}\) and inclined at \(60^{\circ}\) with each other. The acceleration of the body in \(\mathrm{m} / \mathrm{s}^{2}\) is \(\left(\cos 60^{\circ}=0.5\right)\)
371875
A see-saw of length \(6 \mathrm{~m}\) is pivoted at its centre. A child of mass \(20 \mathrm{~kg}\) is sitting at one of its ends. Where should another child, of mass 30 \(\mathrm{kg}\), sit on the other end from the centre of seesaw, so that it is balanced ?
1 \(1 \mathrm{~m}\)
2 \(3 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
C Let, 'n' be the distance of B from the fulcrum. \(\mathrm{F}_{1}=20 \mathrm{~g}\) and \(\mathrm{F}_{2}=30 \mathrm{~g}\) For equilibrium, \(20 \times 3=30 \times x\) \(\mathrm{x}=\frac{60}{30}\) \(\mathrm{x}=2 \mathrm{~m}\)
371872
The maximum and minimum tensions in a string whirling in a circle of radius \(2.5 \mathrm{~m}\) are in the ratio \(5: 3\). Then, its velocity is m.s :
1 7
2 \(\sqrt{98}\)
3 \(\sqrt{4.9}\)
4 \(\sqrt{490}\)
Explanation:
B Given, radius of circle (r) \(=2.5 \mathrm{~m}\) It is assumed that particle although moving in a vertical loop but its speed is constant. Tension of lowest point- \(\mathrm{T}_{\max }=\frac{\mathrm{mv}^{2}}{\mathrm{r}}+\mathrm{mg}\) Tension of highest point- \(\mathrm{T}_{\text {min. }}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}\) Dividing equation (i) by (ii), we get- \(\frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{\frac{\mathrm{mv}}{\mathrm{r}}+\mathrm{mg}}{\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}}=\frac{5}{3} \quad\left[\frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{5}{3} \text { (given) }\right]\) \(\frac{\mathrm{v}^{2}+\mathrm{gr}}{\mathrm{v}^{2}-\mathrm{gr}}=\frac{5}{3}\) \(\mathrm{v}=\sqrt{4 \mathrm{gr}}\) \(\mathrm{v}=\sqrt{4 \times 9.8 \times 2.5}\) \(\mathrm{v}=\sqrt{98} \mathrm{~ms}^{-1}\)
AP EAMCET-25.09.2020
LAWS OF MOTION (ADDITIONAL)
371873
Two bodies of mass \(4 \mathrm{~kg}\) and \(6 \mathrm{~kg}\) are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity \(g\) is
1 \(g / 2\)
2 \(g / 5\)
3 \(g / 10\)
4 \(\mathrm{g}\)
Explanation:
B Given, \(\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=6 \mathrm{~kg}\) For block A - The force balancing equation for \(4 \mathrm{~kg}\) block, \(\mathrm{T}-\mathrm{m}_{1} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}\) \(\mathrm{T}-4 \mathrm{~g}=4 \mathrm{a}\) For block B - Force balancing equation for \(6 \mathrm{~kg}\) block, \(\mathrm{m}_{2} \mathrm{~g}-\mathrm{T}=\mathrm{m}_{2} \mathrm{a}\) \(6 \mathrm{~g}-\mathrm{T}=6 \mathrm{a}\) Adding equation (i) and (ii), we get - \(2 g=10 a\) \(a=g / 5\)
NEET- (Sep.) 2020
LAWS OF MOTION (ADDITIONAL)
371874
A body of mass \(2 \mathrm{~kg}\) is acted upon by two each of magnitude \(1 \mathrm{~N}\) and inclined at \(60^{\circ}\) with each other. The acceleration of the body in \(\mathrm{m} / \mathrm{s}^{2}\) is \(\left(\cos 60^{\circ}=0.5\right)\)
371875
A see-saw of length \(6 \mathrm{~m}\) is pivoted at its centre. A child of mass \(20 \mathrm{~kg}\) is sitting at one of its ends. Where should another child, of mass 30 \(\mathrm{kg}\), sit on the other end from the centre of seesaw, so that it is balanced ?
1 \(1 \mathrm{~m}\)
2 \(3 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
C Let, 'n' be the distance of B from the fulcrum. \(\mathrm{F}_{1}=20 \mathrm{~g}\) and \(\mathrm{F}_{2}=30 \mathrm{~g}\) For equilibrium, \(20 \times 3=30 \times x\) \(\mathrm{x}=\frac{60}{30}\) \(\mathrm{x}=2 \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
LAWS OF MOTION (ADDITIONAL)
371872
The maximum and minimum tensions in a string whirling in a circle of radius \(2.5 \mathrm{~m}\) are in the ratio \(5: 3\). Then, its velocity is m.s :
1 7
2 \(\sqrt{98}\)
3 \(\sqrt{4.9}\)
4 \(\sqrt{490}\)
Explanation:
B Given, radius of circle (r) \(=2.5 \mathrm{~m}\) It is assumed that particle although moving in a vertical loop but its speed is constant. Tension of lowest point- \(\mathrm{T}_{\max }=\frac{\mathrm{mv}^{2}}{\mathrm{r}}+\mathrm{mg}\) Tension of highest point- \(\mathrm{T}_{\text {min. }}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}\) Dividing equation (i) by (ii), we get- \(\frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{\frac{\mathrm{mv}}{\mathrm{r}}+\mathrm{mg}}{\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}}=\frac{5}{3} \quad\left[\frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{5}{3} \text { (given) }\right]\) \(\frac{\mathrm{v}^{2}+\mathrm{gr}}{\mathrm{v}^{2}-\mathrm{gr}}=\frac{5}{3}\) \(\mathrm{v}=\sqrt{4 \mathrm{gr}}\) \(\mathrm{v}=\sqrt{4 \times 9.8 \times 2.5}\) \(\mathrm{v}=\sqrt{98} \mathrm{~ms}^{-1}\)
AP EAMCET-25.09.2020
LAWS OF MOTION (ADDITIONAL)
371873
Two bodies of mass \(4 \mathrm{~kg}\) and \(6 \mathrm{~kg}\) are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity \(g\) is
1 \(g / 2\)
2 \(g / 5\)
3 \(g / 10\)
4 \(\mathrm{g}\)
Explanation:
B Given, \(\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=6 \mathrm{~kg}\) For block A - The force balancing equation for \(4 \mathrm{~kg}\) block, \(\mathrm{T}-\mathrm{m}_{1} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}\) \(\mathrm{T}-4 \mathrm{~g}=4 \mathrm{a}\) For block B - Force balancing equation for \(6 \mathrm{~kg}\) block, \(\mathrm{m}_{2} \mathrm{~g}-\mathrm{T}=\mathrm{m}_{2} \mathrm{a}\) \(6 \mathrm{~g}-\mathrm{T}=6 \mathrm{a}\) Adding equation (i) and (ii), we get - \(2 g=10 a\) \(a=g / 5\)
NEET- (Sep.) 2020
LAWS OF MOTION (ADDITIONAL)
371874
A body of mass \(2 \mathrm{~kg}\) is acted upon by two each of magnitude \(1 \mathrm{~N}\) and inclined at \(60^{\circ}\) with each other. The acceleration of the body in \(\mathrm{m} / \mathrm{s}^{2}\) is \(\left(\cos 60^{\circ}=0.5\right)\)
371875
A see-saw of length \(6 \mathrm{~m}\) is pivoted at its centre. A child of mass \(20 \mathrm{~kg}\) is sitting at one of its ends. Where should another child, of mass 30 \(\mathrm{kg}\), sit on the other end from the centre of seesaw, so that it is balanced ?
1 \(1 \mathrm{~m}\)
2 \(3 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
C Let, 'n' be the distance of B from the fulcrum. \(\mathrm{F}_{1}=20 \mathrm{~g}\) and \(\mathrm{F}_{2}=30 \mathrm{~g}\) For equilibrium, \(20 \times 3=30 \times x\) \(\mathrm{x}=\frac{60}{30}\) \(\mathrm{x}=2 \mathrm{~m}\)