145791
A bullet of mass \(20 \mathrm{~g}\) moving with a velocity of \(200 \mathrm{~m} / \mathrm{s}\) strikes a target and is brought to rest in \(\left(\frac{1}{50}\right)^{\text {th }}\) of a second. The impulse and average force of impact are respectively.
1 \(2 \mathrm{Ns}, 100 \mathrm{~N}\)
2 \(4 \mathrm{Ns}, 200 \mathrm{~N}\)
3 \(2 \mathrm{Ns}, 200 \mathrm{~N}\)
4 \(4 \mathrm{Ns}, 100 \mathrm{~N}\)
Explanation:
B Given, \(\mathrm{m}=20 \mathrm{~g}, \mathrm{v}=200 \mathrm{~m} / \mathrm{s}\) momentum \((\mathrm{p})=\mathrm{mv}\) \(\mathrm{p}=20 \times 10^{-3} \mathrm{~kg} \times 200 \mathrm{~m} / \mathrm{s}\) \(\mathrm{p}=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) Impulse \(=\) change in momentum \((\mathrm{p})=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(\therefore \text { Force } =\frac{\text { Impulse }}{\text { Time }}\) \(\qquad =\frac{4}{\left(\frac{1}{50}\right)}=200 \mathrm{~N}\) The impulse and average force of impact are respectively \(4 \mathrm{Ns}, 200 \mathrm{~N}\).
MHT-CET 2020
Laws of Motion
145792
Let a force \(\overrightarrow{\mathbf{F}}=-\mathbf{F k}\) acts on the origin of Cartesian frame of reference. The moment of force about a point \((1,-1)\) will be
1 \(F(\hat{i}-\hat{j})\)
2 \(F(\hat{i}+\hat{j})\)
3 \(-F(\hat{i}-\hat{j})\)
4 \(-F(\hat{i}+\hat{j})\)
Explanation:
B The torque about the given position, \(\tau=\overrightarrow{\mathrm{r}} \times \mathrm{F}\) Here, \(\quad \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}\) and \(\mathrm{F}=-\mathrm{F} \hat{\mathrm{k}}\) \(\therefore \quad \tau =(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \times(-\mathrm{F} \hat{\mathrm{k}})\) \(=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 0 \\ 0 & 0 & -\mathrm{F}\end{array}\right|\) \(=F \hat{\mathrm{i}}-\hat{\mathrm{j}}(-\mathrm{F})\) \(\tau =\mathrm{F}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\)
MHT-CET 2020
Laws of Motion
145793
A body of mass \(2 \mathrm{~kg}\) is acted upon by two perpendicular forces \(4 \mathrm{~N}\) and \(3 \mathrm{~N}\). The magnitude and direction of the acceleration of the body respectively are.
1 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
2 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
3 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
4 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
Explanation:
A Given that mass of body \((m)=2 \mathrm{~kg}\) two perpendicular forces \(=4 \mathrm{~N}, 3 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{net}}=\sqrt{4^{2}+3^{2}}\) \(=5 \mathrm{~N}\) \(\therefore \mathrm{F}_{\mathrm{net}}=\mathrm{ma}\) \(5=2 . a \Rightarrow a=2.5 \mathrm{~m} / \mathrm{sec}^{2}\) direction of resultant \(\tan \theta=\frac{F_{2}}{F_{1}}\) \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\)
TS EAMCET 28.09.2020
Laws of Motion
145794
The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\) cannot be
1 \(12 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(8 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
B The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\), \(\mathrm{F}_{\max }=10+5\) \(\mathrm{~F}_{\max }=15 \mathrm{~N}\) \(\mathrm{~F}_{\min }=10-5\) \(\mathrm{~F}_{\min }=5 \mathrm{~N}\) Range of force \(=5 \leq \mathrm{F} \leq 15\) Hence, option (b) is correct.
AP EAMCET (Medical)-07.10.2020
Laws of Motion
145796
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
145791
A bullet of mass \(20 \mathrm{~g}\) moving with a velocity of \(200 \mathrm{~m} / \mathrm{s}\) strikes a target and is brought to rest in \(\left(\frac{1}{50}\right)^{\text {th }}\) of a second. The impulse and average force of impact are respectively.
1 \(2 \mathrm{Ns}, 100 \mathrm{~N}\)
2 \(4 \mathrm{Ns}, 200 \mathrm{~N}\)
3 \(2 \mathrm{Ns}, 200 \mathrm{~N}\)
4 \(4 \mathrm{Ns}, 100 \mathrm{~N}\)
Explanation:
B Given, \(\mathrm{m}=20 \mathrm{~g}, \mathrm{v}=200 \mathrm{~m} / \mathrm{s}\) momentum \((\mathrm{p})=\mathrm{mv}\) \(\mathrm{p}=20 \times 10^{-3} \mathrm{~kg} \times 200 \mathrm{~m} / \mathrm{s}\) \(\mathrm{p}=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) Impulse \(=\) change in momentum \((\mathrm{p})=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(\therefore \text { Force } =\frac{\text { Impulse }}{\text { Time }}\) \(\qquad =\frac{4}{\left(\frac{1}{50}\right)}=200 \mathrm{~N}\) The impulse and average force of impact are respectively \(4 \mathrm{Ns}, 200 \mathrm{~N}\).
MHT-CET 2020
Laws of Motion
145792
Let a force \(\overrightarrow{\mathbf{F}}=-\mathbf{F k}\) acts on the origin of Cartesian frame of reference. The moment of force about a point \((1,-1)\) will be
1 \(F(\hat{i}-\hat{j})\)
2 \(F(\hat{i}+\hat{j})\)
3 \(-F(\hat{i}-\hat{j})\)
4 \(-F(\hat{i}+\hat{j})\)
Explanation:
B The torque about the given position, \(\tau=\overrightarrow{\mathrm{r}} \times \mathrm{F}\) Here, \(\quad \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}\) and \(\mathrm{F}=-\mathrm{F} \hat{\mathrm{k}}\) \(\therefore \quad \tau =(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \times(-\mathrm{F} \hat{\mathrm{k}})\) \(=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 0 \\ 0 & 0 & -\mathrm{F}\end{array}\right|\) \(=F \hat{\mathrm{i}}-\hat{\mathrm{j}}(-\mathrm{F})\) \(\tau =\mathrm{F}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\)
MHT-CET 2020
Laws of Motion
145793
A body of mass \(2 \mathrm{~kg}\) is acted upon by two perpendicular forces \(4 \mathrm{~N}\) and \(3 \mathrm{~N}\). The magnitude and direction of the acceleration of the body respectively are.
1 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
2 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
3 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
4 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
Explanation:
A Given that mass of body \((m)=2 \mathrm{~kg}\) two perpendicular forces \(=4 \mathrm{~N}, 3 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{net}}=\sqrt{4^{2}+3^{2}}\) \(=5 \mathrm{~N}\) \(\therefore \mathrm{F}_{\mathrm{net}}=\mathrm{ma}\) \(5=2 . a \Rightarrow a=2.5 \mathrm{~m} / \mathrm{sec}^{2}\) direction of resultant \(\tan \theta=\frac{F_{2}}{F_{1}}\) \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\)
TS EAMCET 28.09.2020
Laws of Motion
145794
The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\) cannot be
1 \(12 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(8 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
B The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\), \(\mathrm{F}_{\max }=10+5\) \(\mathrm{~F}_{\max }=15 \mathrm{~N}\) \(\mathrm{~F}_{\min }=10-5\) \(\mathrm{~F}_{\min }=5 \mathrm{~N}\) Range of force \(=5 \leq \mathrm{F} \leq 15\) Hence, option (b) is correct.
AP EAMCET (Medical)-07.10.2020
Laws of Motion
145796
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
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Laws of Motion
145791
A bullet of mass \(20 \mathrm{~g}\) moving with a velocity of \(200 \mathrm{~m} / \mathrm{s}\) strikes a target and is brought to rest in \(\left(\frac{1}{50}\right)^{\text {th }}\) of a second. The impulse and average force of impact are respectively.
1 \(2 \mathrm{Ns}, 100 \mathrm{~N}\)
2 \(4 \mathrm{Ns}, 200 \mathrm{~N}\)
3 \(2 \mathrm{Ns}, 200 \mathrm{~N}\)
4 \(4 \mathrm{Ns}, 100 \mathrm{~N}\)
Explanation:
B Given, \(\mathrm{m}=20 \mathrm{~g}, \mathrm{v}=200 \mathrm{~m} / \mathrm{s}\) momentum \((\mathrm{p})=\mathrm{mv}\) \(\mathrm{p}=20 \times 10^{-3} \mathrm{~kg} \times 200 \mathrm{~m} / \mathrm{s}\) \(\mathrm{p}=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) Impulse \(=\) change in momentum \((\mathrm{p})=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(\therefore \text { Force } =\frac{\text { Impulse }}{\text { Time }}\) \(\qquad =\frac{4}{\left(\frac{1}{50}\right)}=200 \mathrm{~N}\) The impulse and average force of impact are respectively \(4 \mathrm{Ns}, 200 \mathrm{~N}\).
MHT-CET 2020
Laws of Motion
145792
Let a force \(\overrightarrow{\mathbf{F}}=-\mathbf{F k}\) acts on the origin of Cartesian frame of reference. The moment of force about a point \((1,-1)\) will be
1 \(F(\hat{i}-\hat{j})\)
2 \(F(\hat{i}+\hat{j})\)
3 \(-F(\hat{i}-\hat{j})\)
4 \(-F(\hat{i}+\hat{j})\)
Explanation:
B The torque about the given position, \(\tau=\overrightarrow{\mathrm{r}} \times \mathrm{F}\) Here, \(\quad \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}\) and \(\mathrm{F}=-\mathrm{F} \hat{\mathrm{k}}\) \(\therefore \quad \tau =(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \times(-\mathrm{F} \hat{\mathrm{k}})\) \(=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 0 \\ 0 & 0 & -\mathrm{F}\end{array}\right|\) \(=F \hat{\mathrm{i}}-\hat{\mathrm{j}}(-\mathrm{F})\) \(\tau =\mathrm{F}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\)
MHT-CET 2020
Laws of Motion
145793
A body of mass \(2 \mathrm{~kg}\) is acted upon by two perpendicular forces \(4 \mathrm{~N}\) and \(3 \mathrm{~N}\). The magnitude and direction of the acceleration of the body respectively are.
1 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
2 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
3 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
4 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
Explanation:
A Given that mass of body \((m)=2 \mathrm{~kg}\) two perpendicular forces \(=4 \mathrm{~N}, 3 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{net}}=\sqrt{4^{2}+3^{2}}\) \(=5 \mathrm{~N}\) \(\therefore \mathrm{F}_{\mathrm{net}}=\mathrm{ma}\) \(5=2 . a \Rightarrow a=2.5 \mathrm{~m} / \mathrm{sec}^{2}\) direction of resultant \(\tan \theta=\frac{F_{2}}{F_{1}}\) \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\)
TS EAMCET 28.09.2020
Laws of Motion
145794
The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\) cannot be
1 \(12 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(8 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
B The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\), \(\mathrm{F}_{\max }=10+5\) \(\mathrm{~F}_{\max }=15 \mathrm{~N}\) \(\mathrm{~F}_{\min }=10-5\) \(\mathrm{~F}_{\min }=5 \mathrm{~N}\) Range of force \(=5 \leq \mathrm{F} \leq 15\) Hence, option (b) is correct.
AP EAMCET (Medical)-07.10.2020
Laws of Motion
145796
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
145791
A bullet of mass \(20 \mathrm{~g}\) moving with a velocity of \(200 \mathrm{~m} / \mathrm{s}\) strikes a target and is brought to rest in \(\left(\frac{1}{50}\right)^{\text {th }}\) of a second. The impulse and average force of impact are respectively.
1 \(2 \mathrm{Ns}, 100 \mathrm{~N}\)
2 \(4 \mathrm{Ns}, 200 \mathrm{~N}\)
3 \(2 \mathrm{Ns}, 200 \mathrm{~N}\)
4 \(4 \mathrm{Ns}, 100 \mathrm{~N}\)
Explanation:
B Given, \(\mathrm{m}=20 \mathrm{~g}, \mathrm{v}=200 \mathrm{~m} / \mathrm{s}\) momentum \((\mathrm{p})=\mathrm{mv}\) \(\mathrm{p}=20 \times 10^{-3} \mathrm{~kg} \times 200 \mathrm{~m} / \mathrm{s}\) \(\mathrm{p}=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) Impulse \(=\) change in momentum \((\mathrm{p})=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(\therefore \text { Force } =\frac{\text { Impulse }}{\text { Time }}\) \(\qquad =\frac{4}{\left(\frac{1}{50}\right)}=200 \mathrm{~N}\) The impulse and average force of impact are respectively \(4 \mathrm{Ns}, 200 \mathrm{~N}\).
MHT-CET 2020
Laws of Motion
145792
Let a force \(\overrightarrow{\mathbf{F}}=-\mathbf{F k}\) acts on the origin of Cartesian frame of reference. The moment of force about a point \((1,-1)\) will be
1 \(F(\hat{i}-\hat{j})\)
2 \(F(\hat{i}+\hat{j})\)
3 \(-F(\hat{i}-\hat{j})\)
4 \(-F(\hat{i}+\hat{j})\)
Explanation:
B The torque about the given position, \(\tau=\overrightarrow{\mathrm{r}} \times \mathrm{F}\) Here, \(\quad \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}\) and \(\mathrm{F}=-\mathrm{F} \hat{\mathrm{k}}\) \(\therefore \quad \tau =(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \times(-\mathrm{F} \hat{\mathrm{k}})\) \(=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 0 \\ 0 & 0 & -\mathrm{F}\end{array}\right|\) \(=F \hat{\mathrm{i}}-\hat{\mathrm{j}}(-\mathrm{F})\) \(\tau =\mathrm{F}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\)
MHT-CET 2020
Laws of Motion
145793
A body of mass \(2 \mathrm{~kg}\) is acted upon by two perpendicular forces \(4 \mathrm{~N}\) and \(3 \mathrm{~N}\). The magnitude and direction of the acceleration of the body respectively are.
1 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
2 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
3 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
4 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
Explanation:
A Given that mass of body \((m)=2 \mathrm{~kg}\) two perpendicular forces \(=4 \mathrm{~N}, 3 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{net}}=\sqrt{4^{2}+3^{2}}\) \(=5 \mathrm{~N}\) \(\therefore \mathrm{F}_{\mathrm{net}}=\mathrm{ma}\) \(5=2 . a \Rightarrow a=2.5 \mathrm{~m} / \mathrm{sec}^{2}\) direction of resultant \(\tan \theta=\frac{F_{2}}{F_{1}}\) \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\)
TS EAMCET 28.09.2020
Laws of Motion
145794
The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\) cannot be
1 \(12 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(8 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
B The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\), \(\mathrm{F}_{\max }=10+5\) \(\mathrm{~F}_{\max }=15 \mathrm{~N}\) \(\mathrm{~F}_{\min }=10-5\) \(\mathrm{~F}_{\min }=5 \mathrm{~N}\) Range of force \(=5 \leq \mathrm{F} \leq 15\) Hence, option (b) is correct.
AP EAMCET (Medical)-07.10.2020
Laws of Motion
145796
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
145791
A bullet of mass \(20 \mathrm{~g}\) moving with a velocity of \(200 \mathrm{~m} / \mathrm{s}\) strikes a target and is brought to rest in \(\left(\frac{1}{50}\right)^{\text {th }}\) of a second. The impulse and average force of impact are respectively.
1 \(2 \mathrm{Ns}, 100 \mathrm{~N}\)
2 \(4 \mathrm{Ns}, 200 \mathrm{~N}\)
3 \(2 \mathrm{Ns}, 200 \mathrm{~N}\)
4 \(4 \mathrm{Ns}, 100 \mathrm{~N}\)
Explanation:
B Given, \(\mathrm{m}=20 \mathrm{~g}, \mathrm{v}=200 \mathrm{~m} / \mathrm{s}\) momentum \((\mathrm{p})=\mathrm{mv}\) \(\mathrm{p}=20 \times 10^{-3} \mathrm{~kg} \times 200 \mathrm{~m} / \mathrm{s}\) \(\mathrm{p}=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) Impulse \(=\) change in momentum \((\mathrm{p})=4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(\therefore \text { Force } =\frac{\text { Impulse }}{\text { Time }}\) \(\qquad =\frac{4}{\left(\frac{1}{50}\right)}=200 \mathrm{~N}\) The impulse and average force of impact are respectively \(4 \mathrm{Ns}, 200 \mathrm{~N}\).
MHT-CET 2020
Laws of Motion
145792
Let a force \(\overrightarrow{\mathbf{F}}=-\mathbf{F k}\) acts on the origin of Cartesian frame of reference. The moment of force about a point \((1,-1)\) will be
1 \(F(\hat{i}-\hat{j})\)
2 \(F(\hat{i}+\hat{j})\)
3 \(-F(\hat{i}-\hat{j})\)
4 \(-F(\hat{i}+\hat{j})\)
Explanation:
B The torque about the given position, \(\tau=\overrightarrow{\mathrm{r}} \times \mathrm{F}\) Here, \(\quad \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}\) and \(\mathrm{F}=-\mathrm{F} \hat{\mathrm{k}}\) \(\therefore \quad \tau =(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \times(-\mathrm{F} \hat{\mathrm{k}})\) \(=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 0 \\ 0 & 0 & -\mathrm{F}\end{array}\right|\) \(=F \hat{\mathrm{i}}-\hat{\mathrm{j}}(-\mathrm{F})\) \(\tau =\mathrm{F}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\)
MHT-CET 2020
Laws of Motion
145793
A body of mass \(2 \mathrm{~kg}\) is acted upon by two perpendicular forces \(4 \mathrm{~N}\) and \(3 \mathrm{~N}\). The magnitude and direction of the acceleration of the body respectively are.
1 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
2 \(2.5 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
3 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)\) with respect to the direction of \(4 \mathrm{~N}\) force
4 \(2.0 \mathrm{~m} / \mathrm{s}^{2} \cdot \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)\) with respect to the direction of \(3 \mathrm{~N}\) force
Explanation:
A Given that mass of body \((m)=2 \mathrm{~kg}\) two perpendicular forces \(=4 \mathrm{~N}, 3 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{net}}=\sqrt{4^{2}+3^{2}}\) \(=5 \mathrm{~N}\) \(\therefore \mathrm{F}_{\mathrm{net}}=\mathrm{ma}\) \(5=2 . a \Rightarrow a=2.5 \mathrm{~m} / \mathrm{sec}^{2}\) direction of resultant \(\tan \theta=\frac{F_{2}}{F_{1}}\) \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\)
TS EAMCET 28.09.2020
Laws of Motion
145794
The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\) cannot be
1 \(12 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(8 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
B The resultant force of \(5 \mathrm{~N}\) and \(10 \mathrm{~N}\), \(\mathrm{F}_{\max }=10+5\) \(\mathrm{~F}_{\max }=15 \mathrm{~N}\) \(\mathrm{~F}_{\min }=10-5\) \(\mathrm{~F}_{\min }=5 \mathrm{~N}\) Range of force \(=5 \leq \mathrm{F} \leq 15\) Hence, option (b) is correct.
AP EAMCET (Medical)-07.10.2020
Laws of Motion
145796
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
