NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141202
The speed distance graph is shown below. At what instant of time (in sec) the speed becomes \(4 \mathrm{~m} / \mathrm{s}\) ?
1 \(t=\ln (2)\)
2 \(t=\ln (4)\)
3 \(\mathrm{t}=\ln (8)\)
4 \(t=\ln (6)\)
Explanation:
A From Graph \(v=s+2\) At \(\mathrm{v}=4 \mathrm{~m} / \mathrm{s}, \mathrm{s}=2 \mathrm{~m}\) We know that, \(v=\frac{\mathrm{ds}}{\mathrm{dt}}\) \(\mathrm{v}=\mathrm{s}+2\) \(\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{s}+2\) \(\int \frac{\mathrm{ds}}{\mathrm{s}+2}=\int \mathrm{dt}\) \(\ln (\mathrm{s}+2)=\mathrm{t}+\mathrm{k}\) Given, \(\mathrm{s}=0, \mathrm{t}=0\) \(\mathrm{k}=\ln (2)\) From eqn (1), \(\ln (4)=\mathrm{t}+\ln (2)\) \(\ln (4)-\ln (2)=t\) \(\mathrm{t}=\ln (2)\)
TS EAMCET (Medical) 09.08.2021
Motion in One Dimensions
141203 The acceleration vs distance graph for a particle moving with initial velocity \(5 \mathrm{~m} / \mathrm{s}\) is shown in the figure. The velocity of the particle at \(x=35 \mathrm{~m}\) will be
1 \(20.62 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(25 \mathrm{~m} / \mathrm{s}\)
4 \(50 \mathrm{~m} / \mathrm{s}\)
Explanation:
C We know, Area under the graph \(=\int\) a.ds and \(\quad\) Acceleration \(=\frac{\text { Velocity }}{\text { time }}\) \(\mathrm{a}=\mathrm{V} \cdot \frac{\mathrm{dv}}{\mathrm{ds}}\) Integrating both sides, \(\qquad \int_{\mathrm{S}_{1}}^{\mathrm{S}_{2}} \text { a.ds }=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}\) \(\quad \text { Area under graph }=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}\) \({\left[\frac{1}{2}(5+10) \times 20\right]+[15 \times 10]=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}}\) \({[150+150]=\left[\frac{\mathrm{v}^{2}}{2}\right]_{\mathrm{v}_{1}}^{\mathrm{v}_{2}}}\) \(300=\frac{\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}}{2}\) \(300=\frac{\mathrm{v}_{2}^{2}-(5)^{2}}{2}\) \(600=\mathrm{v}_{2}{ }^{2}-25\) \(\mathrm{v}_{2}=\sqrt{625}\) \(\therefore \mathrm{v}_{2}=25 \mathrm{~m} / \mathrm{s}\)
WB JEE 2021
Motion in One Dimensions
141204
The velocity-time graph of an object is as shown. The displacement during the interval 0 to \(t_{4}\) is
4 area (A) - area (B) + area (C) + area (D)+ area (E)
Explanation:
B Displacement \(=\) Area under the v-t graph. Displacement become positive above the origin line and negative below the line During the interval 0 to \(t_{4}\), Displacement \(\left(0\right.\) to \(\left._{4}\right)=\) Area \((A)-\) Area \((B)+\) Area \((C)\) Area (D)
AP EAMCET-06.09.2021
Motion in One Dimensions
141205
A boat of length \(L\) and mass \(M\) is floating on a stationary lake water. A person of mass \(\mathrm{m}\) walks on the boat from one end to the other. Displacement incurred by the boat with respect to bank of the lake is
1 \(\frac{M}{M-m} L\)
2 \(\frac{m}{M-m} L\)
3 \(\frac{M}{M+m} L\)
4 \(\frac{m}{M+m} L\)
Explanation:
D Given as, Boat length \(=\mathrm{L}\) Mass of man \(=\mathrm{m}\) Mass of boat \(=\mathrm{M}\) Let \(\mathrm{x}=\) displacement of boat in water \(x_{1}(\text { initial })=\frac{m L+M \frac{L}{2}}{m+M}\) \(x_{2}(\text { final })=\frac{m x+M\left(\frac{L}{2}+x\right)}{m+M}\) Now, \(\mathrm{x}_{1}=\mathrm{x}_{2}\) \(\mathrm{mL}+\frac{\mathrm{ML}}{2}=\mathrm{mx}+\frac{\mathrm{ML}}{2}+\mathrm{Mx}\) \(\mathrm{x}=\frac{\mathrm{mL}}{\mathrm{m}+\mathrm{M}}\)
141202
The speed distance graph is shown below. At what instant of time (in sec) the speed becomes \(4 \mathrm{~m} / \mathrm{s}\) ?
1 \(t=\ln (2)\)
2 \(t=\ln (4)\)
3 \(\mathrm{t}=\ln (8)\)
4 \(t=\ln (6)\)
Explanation:
A From Graph \(v=s+2\) At \(\mathrm{v}=4 \mathrm{~m} / \mathrm{s}, \mathrm{s}=2 \mathrm{~m}\) We know that, \(v=\frac{\mathrm{ds}}{\mathrm{dt}}\) \(\mathrm{v}=\mathrm{s}+2\) \(\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{s}+2\) \(\int \frac{\mathrm{ds}}{\mathrm{s}+2}=\int \mathrm{dt}\) \(\ln (\mathrm{s}+2)=\mathrm{t}+\mathrm{k}\) Given, \(\mathrm{s}=0, \mathrm{t}=0\) \(\mathrm{k}=\ln (2)\) From eqn (1), \(\ln (4)=\mathrm{t}+\ln (2)\) \(\ln (4)-\ln (2)=t\) \(\mathrm{t}=\ln (2)\)
TS EAMCET (Medical) 09.08.2021
Motion in One Dimensions
141203 The acceleration vs distance graph for a particle moving with initial velocity \(5 \mathrm{~m} / \mathrm{s}\) is shown in the figure. The velocity of the particle at \(x=35 \mathrm{~m}\) will be
1 \(20.62 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(25 \mathrm{~m} / \mathrm{s}\)
4 \(50 \mathrm{~m} / \mathrm{s}\)
Explanation:
C We know, Area under the graph \(=\int\) a.ds and \(\quad\) Acceleration \(=\frac{\text { Velocity }}{\text { time }}\) \(\mathrm{a}=\mathrm{V} \cdot \frac{\mathrm{dv}}{\mathrm{ds}}\) Integrating both sides, \(\qquad \int_{\mathrm{S}_{1}}^{\mathrm{S}_{2}} \text { a.ds }=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}\) \(\quad \text { Area under graph }=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}\) \({\left[\frac{1}{2}(5+10) \times 20\right]+[15 \times 10]=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}}\) \({[150+150]=\left[\frac{\mathrm{v}^{2}}{2}\right]_{\mathrm{v}_{1}}^{\mathrm{v}_{2}}}\) \(300=\frac{\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}}{2}\) \(300=\frac{\mathrm{v}_{2}^{2}-(5)^{2}}{2}\) \(600=\mathrm{v}_{2}{ }^{2}-25\) \(\mathrm{v}_{2}=\sqrt{625}\) \(\therefore \mathrm{v}_{2}=25 \mathrm{~m} / \mathrm{s}\)
WB JEE 2021
Motion in One Dimensions
141204
The velocity-time graph of an object is as shown. The displacement during the interval 0 to \(t_{4}\) is
4 area (A) - area (B) + area (C) + area (D)+ area (E)
Explanation:
B Displacement \(=\) Area under the v-t graph. Displacement become positive above the origin line and negative below the line During the interval 0 to \(t_{4}\), Displacement \(\left(0\right.\) to \(\left._{4}\right)=\) Area \((A)-\) Area \((B)+\) Area \((C)\) Area (D)
AP EAMCET-06.09.2021
Motion in One Dimensions
141205
A boat of length \(L\) and mass \(M\) is floating on a stationary lake water. A person of mass \(\mathrm{m}\) walks on the boat from one end to the other. Displacement incurred by the boat with respect to bank of the lake is
1 \(\frac{M}{M-m} L\)
2 \(\frac{m}{M-m} L\)
3 \(\frac{M}{M+m} L\)
4 \(\frac{m}{M+m} L\)
Explanation:
D Given as, Boat length \(=\mathrm{L}\) Mass of man \(=\mathrm{m}\) Mass of boat \(=\mathrm{M}\) Let \(\mathrm{x}=\) displacement of boat in water \(x_{1}(\text { initial })=\frac{m L+M \frac{L}{2}}{m+M}\) \(x_{2}(\text { final })=\frac{m x+M\left(\frac{L}{2}+x\right)}{m+M}\) Now, \(\mathrm{x}_{1}=\mathrm{x}_{2}\) \(\mathrm{mL}+\frac{\mathrm{ML}}{2}=\mathrm{mx}+\frac{\mathrm{ML}}{2}+\mathrm{Mx}\) \(\mathrm{x}=\frac{\mathrm{mL}}{\mathrm{m}+\mathrm{M}}\)
141202
The speed distance graph is shown below. At what instant of time (in sec) the speed becomes \(4 \mathrm{~m} / \mathrm{s}\) ?
1 \(t=\ln (2)\)
2 \(t=\ln (4)\)
3 \(\mathrm{t}=\ln (8)\)
4 \(t=\ln (6)\)
Explanation:
A From Graph \(v=s+2\) At \(\mathrm{v}=4 \mathrm{~m} / \mathrm{s}, \mathrm{s}=2 \mathrm{~m}\) We know that, \(v=\frac{\mathrm{ds}}{\mathrm{dt}}\) \(\mathrm{v}=\mathrm{s}+2\) \(\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{s}+2\) \(\int \frac{\mathrm{ds}}{\mathrm{s}+2}=\int \mathrm{dt}\) \(\ln (\mathrm{s}+2)=\mathrm{t}+\mathrm{k}\) Given, \(\mathrm{s}=0, \mathrm{t}=0\) \(\mathrm{k}=\ln (2)\) From eqn (1), \(\ln (4)=\mathrm{t}+\ln (2)\) \(\ln (4)-\ln (2)=t\) \(\mathrm{t}=\ln (2)\)
TS EAMCET (Medical) 09.08.2021
Motion in One Dimensions
141203 The acceleration vs distance graph for a particle moving with initial velocity \(5 \mathrm{~m} / \mathrm{s}\) is shown in the figure. The velocity of the particle at \(x=35 \mathrm{~m}\) will be
1 \(20.62 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(25 \mathrm{~m} / \mathrm{s}\)
4 \(50 \mathrm{~m} / \mathrm{s}\)
Explanation:
C We know, Area under the graph \(=\int\) a.ds and \(\quad\) Acceleration \(=\frac{\text { Velocity }}{\text { time }}\) \(\mathrm{a}=\mathrm{V} \cdot \frac{\mathrm{dv}}{\mathrm{ds}}\) Integrating both sides, \(\qquad \int_{\mathrm{S}_{1}}^{\mathrm{S}_{2}} \text { a.ds }=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}\) \(\quad \text { Area under graph }=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}\) \({\left[\frac{1}{2}(5+10) \times 20\right]+[15 \times 10]=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}}\) \({[150+150]=\left[\frac{\mathrm{v}^{2}}{2}\right]_{\mathrm{v}_{1}}^{\mathrm{v}_{2}}}\) \(300=\frac{\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}}{2}\) \(300=\frac{\mathrm{v}_{2}^{2}-(5)^{2}}{2}\) \(600=\mathrm{v}_{2}{ }^{2}-25\) \(\mathrm{v}_{2}=\sqrt{625}\) \(\therefore \mathrm{v}_{2}=25 \mathrm{~m} / \mathrm{s}\)
WB JEE 2021
Motion in One Dimensions
141204
The velocity-time graph of an object is as shown. The displacement during the interval 0 to \(t_{4}\) is
4 area (A) - area (B) + area (C) + area (D)+ area (E)
Explanation:
B Displacement \(=\) Area under the v-t graph. Displacement become positive above the origin line and negative below the line During the interval 0 to \(t_{4}\), Displacement \(\left(0\right.\) to \(\left._{4}\right)=\) Area \((A)-\) Area \((B)+\) Area \((C)\) Area (D)
AP EAMCET-06.09.2021
Motion in One Dimensions
141205
A boat of length \(L\) and mass \(M\) is floating on a stationary lake water. A person of mass \(\mathrm{m}\) walks on the boat from one end to the other. Displacement incurred by the boat with respect to bank of the lake is
1 \(\frac{M}{M-m} L\)
2 \(\frac{m}{M-m} L\)
3 \(\frac{M}{M+m} L\)
4 \(\frac{m}{M+m} L\)
Explanation:
D Given as, Boat length \(=\mathrm{L}\) Mass of man \(=\mathrm{m}\) Mass of boat \(=\mathrm{M}\) Let \(\mathrm{x}=\) displacement of boat in water \(x_{1}(\text { initial })=\frac{m L+M \frac{L}{2}}{m+M}\) \(x_{2}(\text { final })=\frac{m x+M\left(\frac{L}{2}+x\right)}{m+M}\) Now, \(\mathrm{x}_{1}=\mathrm{x}_{2}\) \(\mathrm{mL}+\frac{\mathrm{ML}}{2}=\mathrm{mx}+\frac{\mathrm{ML}}{2}+\mathrm{Mx}\) \(\mathrm{x}=\frac{\mathrm{mL}}{\mathrm{m}+\mathrm{M}}\)
141202
The speed distance graph is shown below. At what instant of time (in sec) the speed becomes \(4 \mathrm{~m} / \mathrm{s}\) ?
1 \(t=\ln (2)\)
2 \(t=\ln (4)\)
3 \(\mathrm{t}=\ln (8)\)
4 \(t=\ln (6)\)
Explanation:
A From Graph \(v=s+2\) At \(\mathrm{v}=4 \mathrm{~m} / \mathrm{s}, \mathrm{s}=2 \mathrm{~m}\) We know that, \(v=\frac{\mathrm{ds}}{\mathrm{dt}}\) \(\mathrm{v}=\mathrm{s}+2\) \(\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{s}+2\) \(\int \frac{\mathrm{ds}}{\mathrm{s}+2}=\int \mathrm{dt}\) \(\ln (\mathrm{s}+2)=\mathrm{t}+\mathrm{k}\) Given, \(\mathrm{s}=0, \mathrm{t}=0\) \(\mathrm{k}=\ln (2)\) From eqn (1), \(\ln (4)=\mathrm{t}+\ln (2)\) \(\ln (4)-\ln (2)=t\) \(\mathrm{t}=\ln (2)\)
TS EAMCET (Medical) 09.08.2021
Motion in One Dimensions
141203 The acceleration vs distance graph for a particle moving with initial velocity \(5 \mathrm{~m} / \mathrm{s}\) is shown in the figure. The velocity of the particle at \(x=35 \mathrm{~m}\) will be
1 \(20.62 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(25 \mathrm{~m} / \mathrm{s}\)
4 \(50 \mathrm{~m} / \mathrm{s}\)
Explanation:
C We know, Area under the graph \(=\int\) a.ds and \(\quad\) Acceleration \(=\frac{\text { Velocity }}{\text { time }}\) \(\mathrm{a}=\mathrm{V} \cdot \frac{\mathrm{dv}}{\mathrm{ds}}\) Integrating both sides, \(\qquad \int_{\mathrm{S}_{1}}^{\mathrm{S}_{2}} \text { a.ds }=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}\) \(\quad \text { Area under graph }=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}\) \({\left[\frac{1}{2}(5+10) \times 20\right]+[15 \times 10]=\int_{\mathrm{v}_{1}}^{\mathrm{v}_{2}} \mathrm{v} \cdot \mathrm{dv}}\) \({[150+150]=\left[\frac{\mathrm{v}^{2}}{2}\right]_{\mathrm{v}_{1}}^{\mathrm{v}_{2}}}\) \(300=\frac{\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}}{2}\) \(300=\frac{\mathrm{v}_{2}^{2}-(5)^{2}}{2}\) \(600=\mathrm{v}_{2}{ }^{2}-25\) \(\mathrm{v}_{2}=\sqrt{625}\) \(\therefore \mathrm{v}_{2}=25 \mathrm{~m} / \mathrm{s}\)
WB JEE 2021
Motion in One Dimensions
141204
The velocity-time graph of an object is as shown. The displacement during the interval 0 to \(t_{4}\) is
4 area (A) - area (B) + area (C) + area (D)+ area (E)
Explanation:
B Displacement \(=\) Area under the v-t graph. Displacement become positive above the origin line and negative below the line During the interval 0 to \(t_{4}\), Displacement \(\left(0\right.\) to \(\left._{4}\right)=\) Area \((A)-\) Area \((B)+\) Area \((C)\) Area (D)
AP EAMCET-06.09.2021
Motion in One Dimensions
141205
A boat of length \(L\) and mass \(M\) is floating on a stationary lake water. A person of mass \(\mathrm{m}\) walks on the boat from one end to the other. Displacement incurred by the boat with respect to bank of the lake is
1 \(\frac{M}{M-m} L\)
2 \(\frac{m}{M-m} L\)
3 \(\frac{M}{M+m} L\)
4 \(\frac{m}{M+m} L\)
Explanation:
D Given as, Boat length \(=\mathrm{L}\) Mass of man \(=\mathrm{m}\) Mass of boat \(=\mathrm{M}\) Let \(\mathrm{x}=\) displacement of boat in water \(x_{1}(\text { initial })=\frac{m L+M \frac{L}{2}}{m+M}\) \(x_{2}(\text { final })=\frac{m x+M\left(\frac{L}{2}+x\right)}{m+M}\) Now, \(\mathrm{x}_{1}=\mathrm{x}_{2}\) \(\mathrm{mL}+\frac{\mathrm{ML}}{2}=\mathrm{mx}+\frac{\mathrm{ML}}{2}+\mathrm{Mx}\) \(\mathrm{x}=\frac{\mathrm{mL}}{\mathrm{m}+\mathrm{M}}\)
