355817
A uniform rope of linear density \(d\) and length \(l\) is hanging from the edge of a table. The work done in pulling the rope on the table is
1 \(d g l^{2}\)
2 \(\dfrac{d g l}{2}\)
3 \(\dfrac{d g l^{2}}{2}\)
4 \(d^{2} g l\)
Explanation:
Here, mass of rope \(m=l \times d\) For pulling the rope on the table, Distance of centre of gravity moved \(=l / 2\) \(\therefore\) Work done \(=F \times \dfrac{l}{2}=l d g \times \dfrac{l}{2}=\dfrac{d g l^{2}}{2}\)
PHXI06:WORK ENERGY AND POWER
355818
A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on the table is
1 \(MgL/18\)
2 \(MgL\)
3 \(MgL/9\)
4 \(MgL/3\)
Explanation:
The weight of hanging part \(\left(\dfrac{L}{3}\right)\) of chain is \(\left(\dfrac{1}{3} M g\right)\). This weight acts at centre of gravity of the hanging part which is at a depth of \(L\)/6 from the table. As Work done \(=\) force \(\times\) distance \(\therefore W=\dfrac{M g}{3} \times \dfrac{L}{6}=\dfrac{M g L}{18}\)
PHXI06:WORK ENERGY AND POWER
355819
The work done by applied variable force, \(F=x+x^{3}\) from \(x=0 m\) to \(x=2 m\), where \(x\) is displacement, is
1 \(6 J\)
2 \(8 J\)
3 \(10 J\)
4 \(12\,J\)
Explanation:
Given, \(F=x+x^{3}\) and \(x=0\) to \(x=2\) We know that, work done, \(W=\int F d x\) \(\Rightarrow \quad W=\int_{0}^{2}\left(x+x^{3}\right) d x\) \(\Rightarrow W=\left[\dfrac{x^{2}}{2}+\dfrac{x^{4}}{4}\right]_{0}^{2}\) \(W=\left[\dfrac{4}{2}+\dfrac{16}{4}\right]\) \(W=6 J\)
PHXI06:WORK ENERGY AND POWER
355820
A force \(F = 20 + 10y\) acts on a particle in \(y\) direction where\(F\) is in newton and \(y\) in meter. Work done by this force to move the particle from \(y = 0\) to \(y = 1\) \(m\) is
1 30 \(J\)
2 5 \(J\)
3 25 \(J\)
4 20 \(J\)
Explanation:
Work done by variable force is \(W=\int_{y_{i}}^{y_{f}} F d y\) Here, \(y_{i}=0, y_{f}=1 \mathrm{~m}\) \(\therefore \quad W=\int_{0}^{1}(20+10 y) d y=\left[20 y+\dfrac{10 y^{2}}{2}\right]_{0}^{1}=25 J\)
355817
A uniform rope of linear density \(d\) and length \(l\) is hanging from the edge of a table. The work done in pulling the rope on the table is
1 \(d g l^{2}\)
2 \(\dfrac{d g l}{2}\)
3 \(\dfrac{d g l^{2}}{2}\)
4 \(d^{2} g l\)
Explanation:
Here, mass of rope \(m=l \times d\) For pulling the rope on the table, Distance of centre of gravity moved \(=l / 2\) \(\therefore\) Work done \(=F \times \dfrac{l}{2}=l d g \times \dfrac{l}{2}=\dfrac{d g l^{2}}{2}\)
PHXI06:WORK ENERGY AND POWER
355818
A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on the table is
1 \(MgL/18\)
2 \(MgL\)
3 \(MgL/9\)
4 \(MgL/3\)
Explanation:
The weight of hanging part \(\left(\dfrac{L}{3}\right)\) of chain is \(\left(\dfrac{1}{3} M g\right)\). This weight acts at centre of gravity of the hanging part which is at a depth of \(L\)/6 from the table. As Work done \(=\) force \(\times\) distance \(\therefore W=\dfrac{M g}{3} \times \dfrac{L}{6}=\dfrac{M g L}{18}\)
PHXI06:WORK ENERGY AND POWER
355819
The work done by applied variable force, \(F=x+x^{3}\) from \(x=0 m\) to \(x=2 m\), where \(x\) is displacement, is
1 \(6 J\)
2 \(8 J\)
3 \(10 J\)
4 \(12\,J\)
Explanation:
Given, \(F=x+x^{3}\) and \(x=0\) to \(x=2\) We know that, work done, \(W=\int F d x\) \(\Rightarrow \quad W=\int_{0}^{2}\left(x+x^{3}\right) d x\) \(\Rightarrow W=\left[\dfrac{x^{2}}{2}+\dfrac{x^{4}}{4}\right]_{0}^{2}\) \(W=\left[\dfrac{4}{2}+\dfrac{16}{4}\right]\) \(W=6 J\)
PHXI06:WORK ENERGY AND POWER
355820
A force \(F = 20 + 10y\) acts on a particle in \(y\) direction where\(F\) is in newton and \(y\) in meter. Work done by this force to move the particle from \(y = 0\) to \(y = 1\) \(m\) is
1 30 \(J\)
2 5 \(J\)
3 25 \(J\)
4 20 \(J\)
Explanation:
Work done by variable force is \(W=\int_{y_{i}}^{y_{f}} F d y\) Here, \(y_{i}=0, y_{f}=1 \mathrm{~m}\) \(\therefore \quad W=\int_{0}^{1}(20+10 y) d y=\left[20 y+\dfrac{10 y^{2}}{2}\right]_{0}^{1}=25 J\)
355817
A uniform rope of linear density \(d\) and length \(l\) is hanging from the edge of a table. The work done in pulling the rope on the table is
1 \(d g l^{2}\)
2 \(\dfrac{d g l}{2}\)
3 \(\dfrac{d g l^{2}}{2}\)
4 \(d^{2} g l\)
Explanation:
Here, mass of rope \(m=l \times d\) For pulling the rope on the table, Distance of centre of gravity moved \(=l / 2\) \(\therefore\) Work done \(=F \times \dfrac{l}{2}=l d g \times \dfrac{l}{2}=\dfrac{d g l^{2}}{2}\)
PHXI06:WORK ENERGY AND POWER
355818
A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on the table is
1 \(MgL/18\)
2 \(MgL\)
3 \(MgL/9\)
4 \(MgL/3\)
Explanation:
The weight of hanging part \(\left(\dfrac{L}{3}\right)\) of chain is \(\left(\dfrac{1}{3} M g\right)\). This weight acts at centre of gravity of the hanging part which is at a depth of \(L\)/6 from the table. As Work done \(=\) force \(\times\) distance \(\therefore W=\dfrac{M g}{3} \times \dfrac{L}{6}=\dfrac{M g L}{18}\)
PHXI06:WORK ENERGY AND POWER
355819
The work done by applied variable force, \(F=x+x^{3}\) from \(x=0 m\) to \(x=2 m\), where \(x\) is displacement, is
1 \(6 J\)
2 \(8 J\)
3 \(10 J\)
4 \(12\,J\)
Explanation:
Given, \(F=x+x^{3}\) and \(x=0\) to \(x=2\) We know that, work done, \(W=\int F d x\) \(\Rightarrow \quad W=\int_{0}^{2}\left(x+x^{3}\right) d x\) \(\Rightarrow W=\left[\dfrac{x^{2}}{2}+\dfrac{x^{4}}{4}\right]_{0}^{2}\) \(W=\left[\dfrac{4}{2}+\dfrac{16}{4}\right]\) \(W=6 J\)
PHXI06:WORK ENERGY AND POWER
355820
A force \(F = 20 + 10y\) acts on a particle in \(y\) direction where\(F\) is in newton and \(y\) in meter. Work done by this force to move the particle from \(y = 0\) to \(y = 1\) \(m\) is
1 30 \(J\)
2 5 \(J\)
3 25 \(J\)
4 20 \(J\)
Explanation:
Work done by variable force is \(W=\int_{y_{i}}^{y_{f}} F d y\) Here, \(y_{i}=0, y_{f}=1 \mathrm{~m}\) \(\therefore \quad W=\int_{0}^{1}(20+10 y) d y=\left[20 y+\dfrac{10 y^{2}}{2}\right]_{0}^{1}=25 J\)
355817
A uniform rope of linear density \(d\) and length \(l\) is hanging from the edge of a table. The work done in pulling the rope on the table is
1 \(d g l^{2}\)
2 \(\dfrac{d g l}{2}\)
3 \(\dfrac{d g l^{2}}{2}\)
4 \(d^{2} g l\)
Explanation:
Here, mass of rope \(m=l \times d\) For pulling the rope on the table, Distance of centre of gravity moved \(=l / 2\) \(\therefore\) Work done \(=F \times \dfrac{l}{2}=l d g \times \dfrac{l}{2}=\dfrac{d g l^{2}}{2}\)
PHXI06:WORK ENERGY AND POWER
355818
A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on the table is
1 \(MgL/18\)
2 \(MgL\)
3 \(MgL/9\)
4 \(MgL/3\)
Explanation:
The weight of hanging part \(\left(\dfrac{L}{3}\right)\) of chain is \(\left(\dfrac{1}{3} M g\right)\). This weight acts at centre of gravity of the hanging part which is at a depth of \(L\)/6 from the table. As Work done \(=\) force \(\times\) distance \(\therefore W=\dfrac{M g}{3} \times \dfrac{L}{6}=\dfrac{M g L}{18}\)
PHXI06:WORK ENERGY AND POWER
355819
The work done by applied variable force, \(F=x+x^{3}\) from \(x=0 m\) to \(x=2 m\), where \(x\) is displacement, is
1 \(6 J\)
2 \(8 J\)
3 \(10 J\)
4 \(12\,J\)
Explanation:
Given, \(F=x+x^{3}\) and \(x=0\) to \(x=2\) We know that, work done, \(W=\int F d x\) \(\Rightarrow \quad W=\int_{0}^{2}\left(x+x^{3}\right) d x\) \(\Rightarrow W=\left[\dfrac{x^{2}}{2}+\dfrac{x^{4}}{4}\right]_{0}^{2}\) \(W=\left[\dfrac{4}{2}+\dfrac{16}{4}\right]\) \(W=6 J\)
PHXI06:WORK ENERGY AND POWER
355820
A force \(F = 20 + 10y\) acts on a particle in \(y\) direction where\(F\) is in newton and \(y\) in meter. Work done by this force to move the particle from \(y = 0\) to \(y = 1\) \(m\) is
1 30 \(J\)
2 5 \(J\)
3 25 \(J\)
4 20 \(J\)
Explanation:
Work done by variable force is \(W=\int_{y_{i}}^{y_{f}} F d y\) Here, \(y_{i}=0, y_{f}=1 \mathrm{~m}\) \(\therefore \quad W=\int_{0}^{1}(20+10 y) d y=\left[20 y+\dfrac{10 y^{2}}{2}\right]_{0}^{1}=25 J\)
