162567
A body is onithy at rest. It undergoes one dimensions ingtion with constant acceleration. The Power delerved to it at time \(t\) is proportional
1 \(t^{1 / 2}\)
2 \(\mathrm{t}\)
3 \(t^{3 / 2}\)
4 \(\mathrm{t}^2\)
Explanation:
NCERT-I-90
3 RBTS PAPER
162568
Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system
1 neither the mechanical energy nor the linear momentum is conserved
2 both the mechanical energy and the linear momentum are conserved
3 mechanical energy is conserved, but not the linear momentum
4 linear momentum is conserved, but not the mechanical energy
Explanation:
The collision is inelastic. The only momentum is conserved and not the mechanical energy. There is the loss of K.E. in such a collision.
NCERT-I-84
3 RBTS PAPER
162569
A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled \(3 \mathrm{~m}\) is :
1 \(6.5 \mathrm{~J}\)
2 \(2.5 \mathrm{~J}\)
3 \(4 \mathrm{~J}\)
4 \(5 \mathrm{~J}\).
Explanation:
According to work energy theorem Work done by force on the particle \(=\) change in \(\mathrm{KE}\) Work done \(=\) Area under F-x graph \( \int \mathrm{F} . \mathrm{dx}=2 \times 2+\frac{(2+3) \times 1}{2} \) \( \mathrm{~W}=\mathrm{KE}_{\text {final }}-\mathrm{KE}_{\text {initial }}=6.5 \) \( \mathrm{KE}_{\text {initial }}=0 \) \( \mathrm{KE}_{\text {final }}=6.5 \mathrm{~J} . \)
NCERT-I-76
3 RBTS PAPER
162570
Consider a force vector \(\vec{F}=-x \hat{i}+y \hat{j}\). The work done by this force in moving a particle from point \(A(1,0)\) to \(B(0,1)\) along the line segment is :
1 2
2 1
3 \(3 / 2\)
4 \(1 / 2\)
Explanation:
\( W & =\int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d \vec{r} \) \( W & =\int_1^0-x d x+\int_0^1 y d y \Rightarrow W=\left.\frac{-x^2}{2}\right|_1 ^0+\left.\frac{y^2}{2}\right|_0 ^1 \) \( =-\left(\frac{0^2}{2}-\frac{1^2}{2}\right)+\left(\frac{1^2}{2}-\frac{0^2}{2}\right) \Rightarrow W=1 \mathrm{~J} \)
162567
A body is onithy at rest. It undergoes one dimensions ingtion with constant acceleration. The Power delerved to it at time \(t\) is proportional
1 \(t^{1 / 2}\)
2 \(\mathrm{t}\)
3 \(t^{3 / 2}\)
4 \(\mathrm{t}^2\)
Explanation:
NCERT-I-90
3 RBTS PAPER
162568
Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system
1 neither the mechanical energy nor the linear momentum is conserved
2 both the mechanical energy and the linear momentum are conserved
3 mechanical energy is conserved, but not the linear momentum
4 linear momentum is conserved, but not the mechanical energy
Explanation:
The collision is inelastic. The only momentum is conserved and not the mechanical energy. There is the loss of K.E. in such a collision.
NCERT-I-84
3 RBTS PAPER
162569
A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled \(3 \mathrm{~m}\) is :
1 \(6.5 \mathrm{~J}\)
2 \(2.5 \mathrm{~J}\)
3 \(4 \mathrm{~J}\)
4 \(5 \mathrm{~J}\).
Explanation:
According to work energy theorem Work done by force on the particle \(=\) change in \(\mathrm{KE}\) Work done \(=\) Area under F-x graph \( \int \mathrm{F} . \mathrm{dx}=2 \times 2+\frac{(2+3) \times 1}{2} \) \( \mathrm{~W}=\mathrm{KE}_{\text {final }}-\mathrm{KE}_{\text {initial }}=6.5 \) \( \mathrm{KE}_{\text {initial }}=0 \) \( \mathrm{KE}_{\text {final }}=6.5 \mathrm{~J} . \)
NCERT-I-76
3 RBTS PAPER
162570
Consider a force vector \(\vec{F}=-x \hat{i}+y \hat{j}\). The work done by this force in moving a particle from point \(A(1,0)\) to \(B(0,1)\) along the line segment is :
1 2
2 1
3 \(3 / 2\)
4 \(1 / 2\)
Explanation:
\( W & =\int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d \vec{r} \) \( W & =\int_1^0-x d x+\int_0^1 y d y \Rightarrow W=\left.\frac{-x^2}{2}\right|_1 ^0+\left.\frac{y^2}{2}\right|_0 ^1 \) \( =-\left(\frac{0^2}{2}-\frac{1^2}{2}\right)+\left(\frac{1^2}{2}-\frac{0^2}{2}\right) \Rightarrow W=1 \mathrm{~J} \)
162567
A body is onithy at rest. It undergoes one dimensions ingtion with constant acceleration. The Power delerved to it at time \(t\) is proportional
1 \(t^{1 / 2}\)
2 \(\mathrm{t}\)
3 \(t^{3 / 2}\)
4 \(\mathrm{t}^2\)
Explanation:
NCERT-I-90
3 RBTS PAPER
162568
Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system
1 neither the mechanical energy nor the linear momentum is conserved
2 both the mechanical energy and the linear momentum are conserved
3 mechanical energy is conserved, but not the linear momentum
4 linear momentum is conserved, but not the mechanical energy
Explanation:
The collision is inelastic. The only momentum is conserved and not the mechanical energy. There is the loss of K.E. in such a collision.
NCERT-I-84
3 RBTS PAPER
162569
A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled \(3 \mathrm{~m}\) is :
1 \(6.5 \mathrm{~J}\)
2 \(2.5 \mathrm{~J}\)
3 \(4 \mathrm{~J}\)
4 \(5 \mathrm{~J}\).
Explanation:
According to work energy theorem Work done by force on the particle \(=\) change in \(\mathrm{KE}\) Work done \(=\) Area under F-x graph \( \int \mathrm{F} . \mathrm{dx}=2 \times 2+\frac{(2+3) \times 1}{2} \) \( \mathrm{~W}=\mathrm{KE}_{\text {final }}-\mathrm{KE}_{\text {initial }}=6.5 \) \( \mathrm{KE}_{\text {initial }}=0 \) \( \mathrm{KE}_{\text {final }}=6.5 \mathrm{~J} . \)
NCERT-I-76
3 RBTS PAPER
162570
Consider a force vector \(\vec{F}=-x \hat{i}+y \hat{j}\). The work done by this force in moving a particle from point \(A(1,0)\) to \(B(0,1)\) along the line segment is :
1 2
2 1
3 \(3 / 2\)
4 \(1 / 2\)
Explanation:
\( W & =\int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d \vec{r} \) \( W & =\int_1^0-x d x+\int_0^1 y d y \Rightarrow W=\left.\frac{-x^2}{2}\right|_1 ^0+\left.\frac{y^2}{2}\right|_0 ^1 \) \( =-\left(\frac{0^2}{2}-\frac{1^2}{2}\right)+\left(\frac{1^2}{2}-\frac{0^2}{2}\right) \Rightarrow W=1 \mathrm{~J} \)
162567
A body is onithy at rest. It undergoes one dimensions ingtion with constant acceleration. The Power delerved to it at time \(t\) is proportional
1 \(t^{1 / 2}\)
2 \(\mathrm{t}\)
3 \(t^{3 / 2}\)
4 \(\mathrm{t}^2\)
Explanation:
NCERT-I-90
3 RBTS PAPER
162568
Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system
1 neither the mechanical energy nor the linear momentum is conserved
2 both the mechanical energy and the linear momentum are conserved
3 mechanical energy is conserved, but not the linear momentum
4 linear momentum is conserved, but not the mechanical energy
Explanation:
The collision is inelastic. The only momentum is conserved and not the mechanical energy. There is the loss of K.E. in such a collision.
NCERT-I-84
3 RBTS PAPER
162569
A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled \(3 \mathrm{~m}\) is :
1 \(6.5 \mathrm{~J}\)
2 \(2.5 \mathrm{~J}\)
3 \(4 \mathrm{~J}\)
4 \(5 \mathrm{~J}\).
Explanation:
According to work energy theorem Work done by force on the particle \(=\) change in \(\mathrm{KE}\) Work done \(=\) Area under F-x graph \( \int \mathrm{F} . \mathrm{dx}=2 \times 2+\frac{(2+3) \times 1}{2} \) \( \mathrm{~W}=\mathrm{KE}_{\text {final }}-\mathrm{KE}_{\text {initial }}=6.5 \) \( \mathrm{KE}_{\text {initial }}=0 \) \( \mathrm{KE}_{\text {final }}=6.5 \mathrm{~J} . \)
NCERT-I-76
3 RBTS PAPER
162570
Consider a force vector \(\vec{F}=-x \hat{i}+y \hat{j}\). The work done by this force in moving a particle from point \(A(1,0)\) to \(B(0,1)\) along the line segment is :
1 2
2 1
3 \(3 / 2\)
4 \(1 / 2\)
Explanation:
\( W & =\int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d \vec{r} \) \( W & =\int_1^0-x d x+\int_0^1 y d y \Rightarrow W=\left.\frac{-x^2}{2}\right|_1 ^0+\left.\frac{y^2}{2}\right|_0 ^1 \) \( =-\left(\frac{0^2}{2}-\frac{1^2}{2}\right)+\left(\frac{1^2}{2}-\frac{0^2}{2}\right) \Rightarrow W=1 \mathrm{~J} \)
