145782
Which of the following statement is correct?
1 Electromagnetic force is short ranged
2 Relative strength of gravitational force is higher than that of weak nuclear force
3 Range of the weak nuclear force is smaller than that of strong nuclear force
4 Relative strength of strong nuclear force may or may not be higher than that of electromagnetic force
Explanation:
C Relative strength and range of forces | Force | Approximate \lt br> Relative \lt br> Strengths | Range | | :---: | :---: | :---: | | Gravitational | $10^{-38}$ | $\infty$ | | Electromagnetic | $10^{-2}$ | $\infty$ | | Weak nuclear | $10^{-13}$ | $ \lt 10^{-18} \mathrm{~m}$ | | Strong nuclear | 1 | $ \lt 10^{-15} \mathrm{~m}$ | \(\therefore\) Range of the weak nuclear force is smaller than that of strong nuclear force is correct.
TS EAMCET 05.08.2021
Laws of Motion
145783
Two bodies having masses in the ratio \(2: 3\) fall freely under gravity from heights which are in the ratio \(9: 16\). The ratio of their linear momentum on touching the ground is
1 \(2: 9\)
2 \(3: 16\)
3 \(1: 2\)
4 \(3: 2\)
Explanation:
C Given, \(\mathrm{m}_{1}: \mathrm{m}_{2}=2: 3\) \(\mathrm{~h}_{1}: \mathrm{h}_{2}=9: 16\) Linear momentum, \(\mathrm{P}=\mathrm{mv}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) \(\mathrm{v}=\sqrt{2 \mathrm{gh}} \quad\{\because\) initial velocity, \(\mathrm{u}=0\}\) \(\mathrm{P}=\mathrm{m} \sqrt{2 \mathrm{gh}}\) \(\mathrm{P} \propto \mathrm{m} \sqrt{\mathrm{h}}\) The ratio of their linear momentum, \(\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)=\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right) \cdot \sqrt{\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2}{3} \times \sqrt{\frac{9}{16}}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
AP EAMCET-25.08.2021
Laws of Motion
145784
A force acts on a body of mass \(50 \mathrm{~kg}\), for 10 seconds. When the force stops acting on the body, the body covers \(80 \mathrm{~m}\) in the next 10 seconds. What is the magnitude of the force?
1 \(40 \mathrm{~N}\)
2 \(50 \mathrm{~N}\)
3 \(30 \mathrm{~N}\)
4 \(60 \mathrm{~N}\)
Explanation:
A Given data- Mass of body \((\mathrm{m})=50 \mathrm{~kg}\) Time \((t)=10\) seconds Displacement \((\Delta \mathrm{s})=80 \mathrm{~m}\) Force \((\mathrm{F})=\) ? We know that, \(\text { Velocity }=\frac{\Delta \mathrm{s}}{\mathrm{t}}=\frac{80}{10}=8 \mathrm{~m} / \mathrm{sec}\) Then, acceleration \((\mathrm{a})=\frac{\mathrm{v}}{\mathrm{t}}=\frac{8}{10}\) So, \(\quad\) Force \((\mathrm{F})=\mathrm{ma}\) \(\mathrm{F}=50 \times 0.8\) \(\mathrm{F}=40 \mathrm{~N}\)
AP EAMCET-03.09.2021
Laws of Motion
145785
A horizontal beam is pivoted at 0 as shown in the figure. What should be the value of mass ' \(m\) ' to maintain the beam in horizontal position?
1 \(2 \mathrm{~kg}\)
2 \(1 \mathrm{~kg}\)
3 \(4 \mathrm{~kg}\)
4 \(2.5 \mathrm{~kg}\)
Explanation:
C Take moment about 0, \(\mathrm{m}_{1} \mathrm{~g} \mathrm{x}_{1}=\mathrm{m}_{2} \mathrm{~g} \mathrm{x}_{2}\) \(2 \times \mathrm{g} \times 2=\mathrm{m} \times \mathrm{g} \times 1\) \(\mathrm{~m}=4 \mathrm{~kg}\)
145782
Which of the following statement is correct?
1 Electromagnetic force is short ranged
2 Relative strength of gravitational force is higher than that of weak nuclear force
3 Range of the weak nuclear force is smaller than that of strong nuclear force
4 Relative strength of strong nuclear force may or may not be higher than that of electromagnetic force
Explanation:
C Relative strength and range of forces | Force | Approximate \lt br> Relative \lt br> Strengths | Range | | :---: | :---: | :---: | | Gravitational | $10^{-38}$ | $\infty$ | | Electromagnetic | $10^{-2}$ | $\infty$ | | Weak nuclear | $10^{-13}$ | $ \lt 10^{-18} \mathrm{~m}$ | | Strong nuclear | 1 | $ \lt 10^{-15} \mathrm{~m}$ | \(\therefore\) Range of the weak nuclear force is smaller than that of strong nuclear force is correct.
TS EAMCET 05.08.2021
Laws of Motion
145783
Two bodies having masses in the ratio \(2: 3\) fall freely under gravity from heights which are in the ratio \(9: 16\). The ratio of their linear momentum on touching the ground is
1 \(2: 9\)
2 \(3: 16\)
3 \(1: 2\)
4 \(3: 2\)
Explanation:
C Given, \(\mathrm{m}_{1}: \mathrm{m}_{2}=2: 3\) \(\mathrm{~h}_{1}: \mathrm{h}_{2}=9: 16\) Linear momentum, \(\mathrm{P}=\mathrm{mv}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) \(\mathrm{v}=\sqrt{2 \mathrm{gh}} \quad\{\because\) initial velocity, \(\mathrm{u}=0\}\) \(\mathrm{P}=\mathrm{m} \sqrt{2 \mathrm{gh}}\) \(\mathrm{P} \propto \mathrm{m} \sqrt{\mathrm{h}}\) The ratio of their linear momentum, \(\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)=\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right) \cdot \sqrt{\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2}{3} \times \sqrt{\frac{9}{16}}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
AP EAMCET-25.08.2021
Laws of Motion
145784
A force acts on a body of mass \(50 \mathrm{~kg}\), for 10 seconds. When the force stops acting on the body, the body covers \(80 \mathrm{~m}\) in the next 10 seconds. What is the magnitude of the force?
1 \(40 \mathrm{~N}\)
2 \(50 \mathrm{~N}\)
3 \(30 \mathrm{~N}\)
4 \(60 \mathrm{~N}\)
Explanation:
A Given data- Mass of body \((\mathrm{m})=50 \mathrm{~kg}\) Time \((t)=10\) seconds Displacement \((\Delta \mathrm{s})=80 \mathrm{~m}\) Force \((\mathrm{F})=\) ? We know that, \(\text { Velocity }=\frac{\Delta \mathrm{s}}{\mathrm{t}}=\frac{80}{10}=8 \mathrm{~m} / \mathrm{sec}\) Then, acceleration \((\mathrm{a})=\frac{\mathrm{v}}{\mathrm{t}}=\frac{8}{10}\) So, \(\quad\) Force \((\mathrm{F})=\mathrm{ma}\) \(\mathrm{F}=50 \times 0.8\) \(\mathrm{F}=40 \mathrm{~N}\)
AP EAMCET-03.09.2021
Laws of Motion
145785
A horizontal beam is pivoted at 0 as shown in the figure. What should be the value of mass ' \(m\) ' to maintain the beam in horizontal position?
1 \(2 \mathrm{~kg}\)
2 \(1 \mathrm{~kg}\)
3 \(4 \mathrm{~kg}\)
4 \(2.5 \mathrm{~kg}\)
Explanation:
C Take moment about 0, \(\mathrm{m}_{1} \mathrm{~g} \mathrm{x}_{1}=\mathrm{m}_{2} \mathrm{~g} \mathrm{x}_{2}\) \(2 \times \mathrm{g} \times 2=\mathrm{m} \times \mathrm{g} \times 1\) \(\mathrm{~m}=4 \mathrm{~kg}\)
145782
Which of the following statement is correct?
1 Electromagnetic force is short ranged
2 Relative strength of gravitational force is higher than that of weak nuclear force
3 Range of the weak nuclear force is smaller than that of strong nuclear force
4 Relative strength of strong nuclear force may or may not be higher than that of electromagnetic force
Explanation:
C Relative strength and range of forces | Force | Approximate \lt br> Relative \lt br> Strengths | Range | | :---: | :---: | :---: | | Gravitational | $10^{-38}$ | $\infty$ | | Electromagnetic | $10^{-2}$ | $\infty$ | | Weak nuclear | $10^{-13}$ | $ \lt 10^{-18} \mathrm{~m}$ | | Strong nuclear | 1 | $ \lt 10^{-15} \mathrm{~m}$ | \(\therefore\) Range of the weak nuclear force is smaller than that of strong nuclear force is correct.
TS EAMCET 05.08.2021
Laws of Motion
145783
Two bodies having masses in the ratio \(2: 3\) fall freely under gravity from heights which are in the ratio \(9: 16\). The ratio of their linear momentum on touching the ground is
1 \(2: 9\)
2 \(3: 16\)
3 \(1: 2\)
4 \(3: 2\)
Explanation:
C Given, \(\mathrm{m}_{1}: \mathrm{m}_{2}=2: 3\) \(\mathrm{~h}_{1}: \mathrm{h}_{2}=9: 16\) Linear momentum, \(\mathrm{P}=\mathrm{mv}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) \(\mathrm{v}=\sqrt{2 \mathrm{gh}} \quad\{\because\) initial velocity, \(\mathrm{u}=0\}\) \(\mathrm{P}=\mathrm{m} \sqrt{2 \mathrm{gh}}\) \(\mathrm{P} \propto \mathrm{m} \sqrt{\mathrm{h}}\) The ratio of their linear momentum, \(\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)=\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right) \cdot \sqrt{\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2}{3} \times \sqrt{\frac{9}{16}}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
AP EAMCET-25.08.2021
Laws of Motion
145784
A force acts on a body of mass \(50 \mathrm{~kg}\), for 10 seconds. When the force stops acting on the body, the body covers \(80 \mathrm{~m}\) in the next 10 seconds. What is the magnitude of the force?
1 \(40 \mathrm{~N}\)
2 \(50 \mathrm{~N}\)
3 \(30 \mathrm{~N}\)
4 \(60 \mathrm{~N}\)
Explanation:
A Given data- Mass of body \((\mathrm{m})=50 \mathrm{~kg}\) Time \((t)=10\) seconds Displacement \((\Delta \mathrm{s})=80 \mathrm{~m}\) Force \((\mathrm{F})=\) ? We know that, \(\text { Velocity }=\frac{\Delta \mathrm{s}}{\mathrm{t}}=\frac{80}{10}=8 \mathrm{~m} / \mathrm{sec}\) Then, acceleration \((\mathrm{a})=\frac{\mathrm{v}}{\mathrm{t}}=\frac{8}{10}\) So, \(\quad\) Force \((\mathrm{F})=\mathrm{ma}\) \(\mathrm{F}=50 \times 0.8\) \(\mathrm{F}=40 \mathrm{~N}\)
AP EAMCET-03.09.2021
Laws of Motion
145785
A horizontal beam is pivoted at 0 as shown in the figure. What should be the value of mass ' \(m\) ' to maintain the beam in horizontal position?
1 \(2 \mathrm{~kg}\)
2 \(1 \mathrm{~kg}\)
3 \(4 \mathrm{~kg}\)
4 \(2.5 \mathrm{~kg}\)
Explanation:
C Take moment about 0, \(\mathrm{m}_{1} \mathrm{~g} \mathrm{x}_{1}=\mathrm{m}_{2} \mathrm{~g} \mathrm{x}_{2}\) \(2 \times \mathrm{g} \times 2=\mathrm{m} \times \mathrm{g} \times 1\) \(\mathrm{~m}=4 \mathrm{~kg}\)
145782
Which of the following statement is correct?
1 Electromagnetic force is short ranged
2 Relative strength of gravitational force is higher than that of weak nuclear force
3 Range of the weak nuclear force is smaller than that of strong nuclear force
4 Relative strength of strong nuclear force may or may not be higher than that of electromagnetic force
Explanation:
C Relative strength and range of forces | Force | Approximate \lt br> Relative \lt br> Strengths | Range | | :---: | :---: | :---: | | Gravitational | $10^{-38}$ | $\infty$ | | Electromagnetic | $10^{-2}$ | $\infty$ | | Weak nuclear | $10^{-13}$ | $ \lt 10^{-18} \mathrm{~m}$ | | Strong nuclear | 1 | $ \lt 10^{-15} \mathrm{~m}$ | \(\therefore\) Range of the weak nuclear force is smaller than that of strong nuclear force is correct.
TS EAMCET 05.08.2021
Laws of Motion
145783
Two bodies having masses in the ratio \(2: 3\) fall freely under gravity from heights which are in the ratio \(9: 16\). The ratio of their linear momentum on touching the ground is
1 \(2: 9\)
2 \(3: 16\)
3 \(1: 2\)
4 \(3: 2\)
Explanation:
C Given, \(\mathrm{m}_{1}: \mathrm{m}_{2}=2: 3\) \(\mathrm{~h}_{1}: \mathrm{h}_{2}=9: 16\) Linear momentum, \(\mathrm{P}=\mathrm{mv}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) \(\mathrm{v}=\sqrt{2 \mathrm{gh}} \quad\{\because\) initial velocity, \(\mathrm{u}=0\}\) \(\mathrm{P}=\mathrm{m} \sqrt{2 \mathrm{gh}}\) \(\mathrm{P} \propto \mathrm{m} \sqrt{\mathrm{h}}\) The ratio of their linear momentum, \(\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)=\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right) \cdot \sqrt{\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2}{3} \times \sqrt{\frac{9}{16}}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
AP EAMCET-25.08.2021
Laws of Motion
145784
A force acts on a body of mass \(50 \mathrm{~kg}\), for 10 seconds. When the force stops acting on the body, the body covers \(80 \mathrm{~m}\) in the next 10 seconds. What is the magnitude of the force?
1 \(40 \mathrm{~N}\)
2 \(50 \mathrm{~N}\)
3 \(30 \mathrm{~N}\)
4 \(60 \mathrm{~N}\)
Explanation:
A Given data- Mass of body \((\mathrm{m})=50 \mathrm{~kg}\) Time \((t)=10\) seconds Displacement \((\Delta \mathrm{s})=80 \mathrm{~m}\) Force \((\mathrm{F})=\) ? We know that, \(\text { Velocity }=\frac{\Delta \mathrm{s}}{\mathrm{t}}=\frac{80}{10}=8 \mathrm{~m} / \mathrm{sec}\) Then, acceleration \((\mathrm{a})=\frac{\mathrm{v}}{\mathrm{t}}=\frac{8}{10}\) So, \(\quad\) Force \((\mathrm{F})=\mathrm{ma}\) \(\mathrm{F}=50 \times 0.8\) \(\mathrm{F}=40 \mathrm{~N}\)
AP EAMCET-03.09.2021
Laws of Motion
145785
A horizontal beam is pivoted at 0 as shown in the figure. What should be the value of mass ' \(m\) ' to maintain the beam in horizontal position?
1 \(2 \mathrm{~kg}\)
2 \(1 \mathrm{~kg}\)
3 \(4 \mathrm{~kg}\)
4 \(2.5 \mathrm{~kg}\)
Explanation:
C Take moment about 0, \(\mathrm{m}_{1} \mathrm{~g} \mathrm{x}_{1}=\mathrm{m}_{2} \mathrm{~g} \mathrm{x}_{2}\) \(2 \times \mathrm{g} \times 2=\mathrm{m} \times \mathrm{g} \times 1\) \(\mathrm{~m}=4 \mathrm{~kg}\)
