365701
If linear density of a rod of length \(3 m\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is
1 \(\dfrac{12}{7} m\)
2 \(\dfrac{7}{3} m\)
3 \(\dfrac{9}{7} m\)
4 \(\dfrac{10}{7} m\)
Explanation:
Linear density of the rod varies with distance as \(\begin{aligned}& \lambda=2+x \\& \dfrac{d m}{d x}=\lambda \text { [given] } \therefore d m=\lambda d x\end{aligned}\) Position of centre of mass \({x_{cm}} = \frac{{\int d m \times x}}{{\int d m}}\) \( = \frac{{\int\limits_0^3 {\left( {\lambda dx} \right) \times x} }}{{\int\limits_0^3 {\lambda dx} }} = \frac{{\int\limits_0^3 {\left( {2 + x} \right)} \times xdx}}{{\int\limits_0^3 {\left( {2 + x} \right)} \times dx}}\) \( = \frac{{\left[ {{x^2} + \frac{{{x^3}}}{3}} \right]_0^3}}{{\left[ {2x + \frac{{{x^2}}}{2}} \right]_0^3}} = \frac{{9 + 9}}{{6 + \frac{9}{2}}} = \frac{{12}}{7}m.\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365702
From a uniform disc of radius \(R\), a circular section of radius \(\dfrac{R}{2}\) is cut out. The centre of the hole is at \(\dfrac{R}{2}\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
1 \(\dfrac{R}{3}\) to the right of centre \(O\)
2 \(\dfrac{R}{6}\) to the right of centre \(O\)
3 \(\dfrac{R}{6}\) to the left of centre \(O\)
4 \(\dfrac{R}{3}\) to the left of centre \(O\)
Explanation:
Let mass per unit area of the disc be \(\sigma\). \(\therefore\) Mass of the disc \((M) = \pi {R^2}\sigma \). Mass of the portion removed from the disc \(\left(M^{\prime}\right)\) \(=\pi\left(\dfrac{R}{2}\right)^{2} \sigma=\dfrac{M}{4}\) The position of centre of mass of the remaining portion is \(x = \frac{{ - \frac{M}{4} \times \frac{R}{2}}}{{M - \frac{M}{4}}}\) \( = - \frac{{MR}}{8} \times \frac{4}{{3M}}\) \(x = \frac{{ - R}}{6}\)
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365703
Which of the following statements are correct?
1 Centre of mass of a body always coincides with the centre of gravity of the body.
2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.
3 A couple on a body produces both translational and rotational motion in a body.
4 Mechanical advantage greater than one means that small effort can be used to lift a large load.
Explanation:
Centre of gravity of a body is the point at which the total gravitational torque on body is zero. Centre of mass and centre of gravity coincides only for symmetrical bodies. Hence statements (1) and (2) are incorrect. A couple of a body produces rotational motion only. Hence statement (3) is incorrect. Mechanical advantage greater than one means that the system will require a force that is less than the load in order to move it. Hence only statemnet (4) is correct.
NEET - 2017
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365704
A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity \(G\). When the card board is in equilibrium (translational and rotational) then predict the correct option.
As the card board is in translational equilibrium the net vertical force should be zero. \(\vec{R}=M \vec{g}\) For rotational equilibrium the net torque should be zero about any point. The torque about \(G\) is \(\vec{\tau}_{G}=\sum \vec{r}_{i} \times \vec{F}_{g}=\sum \vec{r}_{i} \times \vec{m}_{i} \vec{g}=0\) Where \(\vec{r}_{i}\) is the position vector of the \(p\) article \(m_{i}\) is the mass of the \(i^{\text {th }}\) particle. \(\vec{g}\) is the acceleration vector. For a small object \(\vec{g}\) is same for all the particles so \(\vec{g}\) can be taken out of summation. \(\vec{g} \sum \vec{r}_{i} m_{i}=0 \Rightarrow \sum m_{i} \vec{r}_{i}=\dfrac{\vec{r}_{C M}}{\sum m_{i}}=0\) So the \(CM\) coincides with centre of gravity So option (4) is correct.
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365705
An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines \(A{A_1},B{B_1}\,\& \,C{C_1}\). Then the centre of gravity of the body lies
1 on line \(A{A_1}\)
2 on line \(B{B_1}\)
3 on line \(C{C_1}\)
4 All the above
Explanation:
Centre of gravity is the intersection point of all the three lines. So it lies on all the three lines. So option (4) is correct.
365701
If linear density of a rod of length \(3 m\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is
1 \(\dfrac{12}{7} m\)
2 \(\dfrac{7}{3} m\)
3 \(\dfrac{9}{7} m\)
4 \(\dfrac{10}{7} m\)
Explanation:
Linear density of the rod varies with distance as \(\begin{aligned}& \lambda=2+x \\& \dfrac{d m}{d x}=\lambda \text { [given] } \therefore d m=\lambda d x\end{aligned}\) Position of centre of mass \({x_{cm}} = \frac{{\int d m \times x}}{{\int d m}}\) \( = \frac{{\int\limits_0^3 {\left( {\lambda dx} \right) \times x} }}{{\int\limits_0^3 {\lambda dx} }} = \frac{{\int\limits_0^3 {\left( {2 + x} \right)} \times xdx}}{{\int\limits_0^3 {\left( {2 + x} \right)} \times dx}}\) \( = \frac{{\left[ {{x^2} + \frac{{{x^3}}}{3}} \right]_0^3}}{{\left[ {2x + \frac{{{x^2}}}{2}} \right]_0^3}} = \frac{{9 + 9}}{{6 + \frac{9}{2}}} = \frac{{12}}{7}m.\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365702
From a uniform disc of radius \(R\), a circular section of radius \(\dfrac{R}{2}\) is cut out. The centre of the hole is at \(\dfrac{R}{2}\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
1 \(\dfrac{R}{3}\) to the right of centre \(O\)
2 \(\dfrac{R}{6}\) to the right of centre \(O\)
3 \(\dfrac{R}{6}\) to the left of centre \(O\)
4 \(\dfrac{R}{3}\) to the left of centre \(O\)
Explanation:
Let mass per unit area of the disc be \(\sigma\). \(\therefore\) Mass of the disc \((M) = \pi {R^2}\sigma \). Mass of the portion removed from the disc \(\left(M^{\prime}\right)\) \(=\pi\left(\dfrac{R}{2}\right)^{2} \sigma=\dfrac{M}{4}\) The position of centre of mass of the remaining portion is \(x = \frac{{ - \frac{M}{4} \times \frac{R}{2}}}{{M - \frac{M}{4}}}\) \( = - \frac{{MR}}{8} \times \frac{4}{{3M}}\) \(x = \frac{{ - R}}{6}\)
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365703
Which of the following statements are correct?
1 Centre of mass of a body always coincides with the centre of gravity of the body.
2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.
3 A couple on a body produces both translational and rotational motion in a body.
4 Mechanical advantage greater than one means that small effort can be used to lift a large load.
Explanation:
Centre of gravity of a body is the point at which the total gravitational torque on body is zero. Centre of mass and centre of gravity coincides only for symmetrical bodies. Hence statements (1) and (2) are incorrect. A couple of a body produces rotational motion only. Hence statement (3) is incorrect. Mechanical advantage greater than one means that the system will require a force that is less than the load in order to move it. Hence only statemnet (4) is correct.
NEET - 2017
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365704
A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity \(G\). When the card board is in equilibrium (translational and rotational) then predict the correct option.
As the card board is in translational equilibrium the net vertical force should be zero. \(\vec{R}=M \vec{g}\) For rotational equilibrium the net torque should be zero about any point. The torque about \(G\) is \(\vec{\tau}_{G}=\sum \vec{r}_{i} \times \vec{F}_{g}=\sum \vec{r}_{i} \times \vec{m}_{i} \vec{g}=0\) Where \(\vec{r}_{i}\) is the position vector of the \(p\) article \(m_{i}\) is the mass of the \(i^{\text {th }}\) particle. \(\vec{g}\) is the acceleration vector. For a small object \(\vec{g}\) is same for all the particles so \(\vec{g}\) can be taken out of summation. \(\vec{g} \sum \vec{r}_{i} m_{i}=0 \Rightarrow \sum m_{i} \vec{r}_{i}=\dfrac{\vec{r}_{C M}}{\sum m_{i}}=0\) So the \(CM\) coincides with centre of gravity So option (4) is correct.
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365705
An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines \(A{A_1},B{B_1}\,\& \,C{C_1}\). Then the centre of gravity of the body lies
1 on line \(A{A_1}\)
2 on line \(B{B_1}\)
3 on line \(C{C_1}\)
4 All the above
Explanation:
Centre of gravity is the intersection point of all the three lines. So it lies on all the three lines. So option (4) is correct.
365701
If linear density of a rod of length \(3 m\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is
1 \(\dfrac{12}{7} m\)
2 \(\dfrac{7}{3} m\)
3 \(\dfrac{9}{7} m\)
4 \(\dfrac{10}{7} m\)
Explanation:
Linear density of the rod varies with distance as \(\begin{aligned}& \lambda=2+x \\& \dfrac{d m}{d x}=\lambda \text { [given] } \therefore d m=\lambda d x\end{aligned}\) Position of centre of mass \({x_{cm}} = \frac{{\int d m \times x}}{{\int d m}}\) \( = \frac{{\int\limits_0^3 {\left( {\lambda dx} \right) \times x} }}{{\int\limits_0^3 {\lambda dx} }} = \frac{{\int\limits_0^3 {\left( {2 + x} \right)} \times xdx}}{{\int\limits_0^3 {\left( {2 + x} \right)} \times dx}}\) \( = \frac{{\left[ {{x^2} + \frac{{{x^3}}}{3}} \right]_0^3}}{{\left[ {2x + \frac{{{x^2}}}{2}} \right]_0^3}} = \frac{{9 + 9}}{{6 + \frac{9}{2}}} = \frac{{12}}{7}m.\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365702
From a uniform disc of radius \(R\), a circular section of radius \(\dfrac{R}{2}\) is cut out. The centre of the hole is at \(\dfrac{R}{2}\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
1 \(\dfrac{R}{3}\) to the right of centre \(O\)
2 \(\dfrac{R}{6}\) to the right of centre \(O\)
3 \(\dfrac{R}{6}\) to the left of centre \(O\)
4 \(\dfrac{R}{3}\) to the left of centre \(O\)
Explanation:
Let mass per unit area of the disc be \(\sigma\). \(\therefore\) Mass of the disc \((M) = \pi {R^2}\sigma \). Mass of the portion removed from the disc \(\left(M^{\prime}\right)\) \(=\pi\left(\dfrac{R}{2}\right)^{2} \sigma=\dfrac{M}{4}\) The position of centre of mass of the remaining portion is \(x = \frac{{ - \frac{M}{4} \times \frac{R}{2}}}{{M - \frac{M}{4}}}\) \( = - \frac{{MR}}{8} \times \frac{4}{{3M}}\) \(x = \frac{{ - R}}{6}\)
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365703
Which of the following statements are correct?
1 Centre of mass of a body always coincides with the centre of gravity of the body.
2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.
3 A couple on a body produces both translational and rotational motion in a body.
4 Mechanical advantage greater than one means that small effort can be used to lift a large load.
Explanation:
Centre of gravity of a body is the point at which the total gravitational torque on body is zero. Centre of mass and centre of gravity coincides only for symmetrical bodies. Hence statements (1) and (2) are incorrect. A couple of a body produces rotational motion only. Hence statement (3) is incorrect. Mechanical advantage greater than one means that the system will require a force that is less than the load in order to move it. Hence only statemnet (4) is correct.
NEET - 2017
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365704
A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity \(G\). When the card board is in equilibrium (translational and rotational) then predict the correct option.
As the card board is in translational equilibrium the net vertical force should be zero. \(\vec{R}=M \vec{g}\) For rotational equilibrium the net torque should be zero about any point. The torque about \(G\) is \(\vec{\tau}_{G}=\sum \vec{r}_{i} \times \vec{F}_{g}=\sum \vec{r}_{i} \times \vec{m}_{i} \vec{g}=0\) Where \(\vec{r}_{i}\) is the position vector of the \(p\) article \(m_{i}\) is the mass of the \(i^{\text {th }}\) particle. \(\vec{g}\) is the acceleration vector. For a small object \(\vec{g}\) is same for all the particles so \(\vec{g}\) can be taken out of summation. \(\vec{g} \sum \vec{r}_{i} m_{i}=0 \Rightarrow \sum m_{i} \vec{r}_{i}=\dfrac{\vec{r}_{C M}}{\sum m_{i}}=0\) So the \(CM\) coincides with centre of gravity So option (4) is correct.
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365705
An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines \(A{A_1},B{B_1}\,\& \,C{C_1}\). Then the centre of gravity of the body lies
1 on line \(A{A_1}\)
2 on line \(B{B_1}\)
3 on line \(C{C_1}\)
4 All the above
Explanation:
Centre of gravity is the intersection point of all the three lines. So it lies on all the three lines. So option (4) is correct.
365701
If linear density of a rod of length \(3 m\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is
1 \(\dfrac{12}{7} m\)
2 \(\dfrac{7}{3} m\)
3 \(\dfrac{9}{7} m\)
4 \(\dfrac{10}{7} m\)
Explanation:
Linear density of the rod varies with distance as \(\begin{aligned}& \lambda=2+x \\& \dfrac{d m}{d x}=\lambda \text { [given] } \therefore d m=\lambda d x\end{aligned}\) Position of centre of mass \({x_{cm}} = \frac{{\int d m \times x}}{{\int d m}}\) \( = \frac{{\int\limits_0^3 {\left( {\lambda dx} \right) \times x} }}{{\int\limits_0^3 {\lambda dx} }} = \frac{{\int\limits_0^3 {\left( {2 + x} \right)} \times xdx}}{{\int\limits_0^3 {\left( {2 + x} \right)} \times dx}}\) \( = \frac{{\left[ {{x^2} + \frac{{{x^3}}}{3}} \right]_0^3}}{{\left[ {2x + \frac{{{x^2}}}{2}} \right]_0^3}} = \frac{{9 + 9}}{{6 + \frac{9}{2}}} = \frac{{12}}{7}m.\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365702
From a uniform disc of radius \(R\), a circular section of radius \(\dfrac{R}{2}\) is cut out. The centre of the hole is at \(\dfrac{R}{2}\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
1 \(\dfrac{R}{3}\) to the right of centre \(O\)
2 \(\dfrac{R}{6}\) to the right of centre \(O\)
3 \(\dfrac{R}{6}\) to the left of centre \(O\)
4 \(\dfrac{R}{3}\) to the left of centre \(O\)
Explanation:
Let mass per unit area of the disc be \(\sigma\). \(\therefore\) Mass of the disc \((M) = \pi {R^2}\sigma \). Mass of the portion removed from the disc \(\left(M^{\prime}\right)\) \(=\pi\left(\dfrac{R}{2}\right)^{2} \sigma=\dfrac{M}{4}\) The position of centre of mass of the remaining portion is \(x = \frac{{ - \frac{M}{4} \times \frac{R}{2}}}{{M - \frac{M}{4}}}\) \( = - \frac{{MR}}{8} \times \frac{4}{{3M}}\) \(x = \frac{{ - R}}{6}\)
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365703
Which of the following statements are correct?
1 Centre of mass of a body always coincides with the centre of gravity of the body.
2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.
3 A couple on a body produces both translational and rotational motion in a body.
4 Mechanical advantage greater than one means that small effort can be used to lift a large load.
Explanation:
Centre of gravity of a body is the point at which the total gravitational torque on body is zero. Centre of mass and centre of gravity coincides only for symmetrical bodies. Hence statements (1) and (2) are incorrect. A couple of a body produces rotational motion only. Hence statement (3) is incorrect. Mechanical advantage greater than one means that the system will require a force that is less than the load in order to move it. Hence only statemnet (4) is correct.
NEET - 2017
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365704
A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity \(G\). When the card board is in equilibrium (translational and rotational) then predict the correct option.
As the card board is in translational equilibrium the net vertical force should be zero. \(\vec{R}=M \vec{g}\) For rotational equilibrium the net torque should be zero about any point. The torque about \(G\) is \(\vec{\tau}_{G}=\sum \vec{r}_{i} \times \vec{F}_{g}=\sum \vec{r}_{i} \times \vec{m}_{i} \vec{g}=0\) Where \(\vec{r}_{i}\) is the position vector of the \(p\) article \(m_{i}\) is the mass of the \(i^{\text {th }}\) particle. \(\vec{g}\) is the acceleration vector. For a small object \(\vec{g}\) is same for all the particles so \(\vec{g}\) can be taken out of summation. \(\vec{g} \sum \vec{r}_{i} m_{i}=0 \Rightarrow \sum m_{i} \vec{r}_{i}=\dfrac{\vec{r}_{C M}}{\sum m_{i}}=0\) So the \(CM\) coincides with centre of gravity So option (4) is correct.
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365705
An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines \(A{A_1},B{B_1}\,\& \,C{C_1}\). Then the centre of gravity of the body lies
1 on line \(A{A_1}\)
2 on line \(B{B_1}\)
3 on line \(C{C_1}\)
4 All the above
Explanation:
Centre of gravity is the intersection point of all the three lines. So it lies on all the three lines. So option (4) is correct.
365701
If linear density of a rod of length \(3 m\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is
1 \(\dfrac{12}{7} m\)
2 \(\dfrac{7}{3} m\)
3 \(\dfrac{9}{7} m\)
4 \(\dfrac{10}{7} m\)
Explanation:
Linear density of the rod varies with distance as \(\begin{aligned}& \lambda=2+x \\& \dfrac{d m}{d x}=\lambda \text { [given] } \therefore d m=\lambda d x\end{aligned}\) Position of centre of mass \({x_{cm}} = \frac{{\int d m \times x}}{{\int d m}}\) \( = \frac{{\int\limits_0^3 {\left( {\lambda dx} \right) \times x} }}{{\int\limits_0^3 {\lambda dx} }} = \frac{{\int\limits_0^3 {\left( {2 + x} \right)} \times xdx}}{{\int\limits_0^3 {\left( {2 + x} \right)} \times dx}}\) \( = \frac{{\left[ {{x^2} + \frac{{{x^3}}}{3}} \right]_0^3}}{{\left[ {2x + \frac{{{x^2}}}{2}} \right]_0^3}} = \frac{{9 + 9}}{{6 + \frac{9}{2}}} = \frac{{12}}{7}m.\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365702
From a uniform disc of radius \(R\), a circular section of radius \(\dfrac{R}{2}\) is cut out. The centre of the hole is at \(\dfrac{R}{2}\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
1 \(\dfrac{R}{3}\) to the right of centre \(O\)
2 \(\dfrac{R}{6}\) to the right of centre \(O\)
3 \(\dfrac{R}{6}\) to the left of centre \(O\)
4 \(\dfrac{R}{3}\) to the left of centre \(O\)
Explanation:
Let mass per unit area of the disc be \(\sigma\). \(\therefore\) Mass of the disc \((M) = \pi {R^2}\sigma \). Mass of the portion removed from the disc \(\left(M^{\prime}\right)\) \(=\pi\left(\dfrac{R}{2}\right)^{2} \sigma=\dfrac{M}{4}\) The position of centre of mass of the remaining portion is \(x = \frac{{ - \frac{M}{4} \times \frac{R}{2}}}{{M - \frac{M}{4}}}\) \( = - \frac{{MR}}{8} \times \frac{4}{{3M}}\) \(x = \frac{{ - R}}{6}\)
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365703
Which of the following statements are correct?
1 Centre of mass of a body always coincides with the centre of gravity of the body.
2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.
3 A couple on a body produces both translational and rotational motion in a body.
4 Mechanical advantage greater than one means that small effort can be used to lift a large load.
Explanation:
Centre of gravity of a body is the point at which the total gravitational torque on body is zero. Centre of mass and centre of gravity coincides only for symmetrical bodies. Hence statements (1) and (2) are incorrect. A couple of a body produces rotational motion only. Hence statement (3) is incorrect. Mechanical advantage greater than one means that the system will require a force that is less than the load in order to move it. Hence only statemnet (4) is correct.
NEET - 2017
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365704
A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity \(G\). When the card board is in equilibrium (translational and rotational) then predict the correct option.
As the card board is in translational equilibrium the net vertical force should be zero. \(\vec{R}=M \vec{g}\) For rotational equilibrium the net torque should be zero about any point. The torque about \(G\) is \(\vec{\tau}_{G}=\sum \vec{r}_{i} \times \vec{F}_{g}=\sum \vec{r}_{i} \times \vec{m}_{i} \vec{g}=0\) Where \(\vec{r}_{i}\) is the position vector of the \(p\) article \(m_{i}\) is the mass of the \(i^{\text {th }}\) particle. \(\vec{g}\) is the acceleration vector. For a small object \(\vec{g}\) is same for all the particles so \(\vec{g}\) can be taken out of summation. \(\vec{g} \sum \vec{r}_{i} m_{i}=0 \Rightarrow \sum m_{i} \vec{r}_{i}=\dfrac{\vec{r}_{C M}}{\sum m_{i}}=0\) So the \(CM\) coincides with centre of gravity So option (4) is correct.
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365705
An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines \(A{A_1},B{B_1}\,\& \,C{C_1}\). Then the centre of gravity of the body lies
1 on line \(A{A_1}\)
2 on line \(B{B_1}\)
3 on line \(C{C_1}\)
4 All the above
Explanation:
Centre of gravity is the intersection point of all the three lines. So it lies on all the three lines. So option (4) is correct.
