Physics Test - 15

Given,

Initial velocity \(=3\) m/s

Time \(=20\) s

Total distance covered by particle,

\(d=v_{1} t_{1}+v_{2} t_{2}+v_{3} t_{3}\)

\(\because t_{1}=t_{2}=t_{3}=20\) s

\(T=3 \times 20=60\) s

\(v=\frac{d}{t}\)

\(\Rightarrow \frac{240}{60}=4\) m/s

Let the spheres collide after time t, when the smaller sphere covered distance \(x_{1}\) and bigger sphere covered distance \(x_{2}\).

The gravitational force acting between two spheres depends on the distance which is a variable quantity.

Just before the collision, total distance travelled by both \(=9 R\)

If \(x\) is the distance travelled by the block of mass \(M\) and \(9 R-x\) the distance travelled by block of mass \(5 \mathrm{M}\).

As the centre of mass of system, not change, then:

\(M x=5 M(9 R-x)\) \(x=7.5 R\)

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

Work done, \((W)=\Delta K=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)

Where \(m\) is the mass of the object, \(v\) is the final velocity of the object and \(u\) is the initial velocity of the object.

Work-energy theorem for a variable force:

Kinetic energy, \(K=\frac{1}{2} m v^{2}\)

\( \frac{d K}{d t}=\frac{d\left(\frac{1}{2} m v^{2}\right)}{d t} \)

\(\Rightarrow \frac{d K}{d t}=m \frac{d v}{d t} v\)

\(\Rightarrow \frac{d K}{d t}=m a v\)

\(\Rightarrow \frac{d K}{d t}=F v \)

\(\Rightarrow \frac{d K}{d t}=F \frac{d x}{d t} \)

\(\Rightarrow d K=F d x\)

On integrating, we get

\( \int_{K_{i}}^{K_{f}} d K=\int_{x_{i}}^{x_{f}} F d x \)

\(\Rightarrow \Delta K=\int_{x_{i}}^{x_{f}} F d x\)

All the given options are correct as they are all different forms of representing the work-energy theorem for a variable force.

Since it is given that the coefficient of cubical expansion of alcohol is more than that of the metal.

Upthrust = (volume of the metal ball)× (density of liquid)×g

With the increase in temperature volume of the ball will increase and the density of the liquid will decrease. But the coefficient of cubical expansion of liquid is more. Hence the second effect is more dominating. Therefore upthrust, at higher temperatures will be less, or apparent weight will be more.

A conical pendulum consists of a mass on the end of a string suspended from a point which moves in a circular path.

Let us consider a conical pendulum having the mass \(m\) revolving in a circle at a constant velocity \(v\) on a string of length \(l\) at an angle of \(\theta\).

There will be two forces acting on the mass, Tension and centripetal force.

The Tension exerted can be resolved into a horizontal component, \(T \sin (\theta)\) and vertical component

\(\operatorname{Tcos}(\theta)\)

The horizontal component of the tension experience centripetal force since the conical pendulum travels in a circular path of radius r with a constant velocity v

\(T \sin \theta=\frac{m v^{2}}{r}\)

We can rearrange the above equation as

\(T=\frac{m v^{2}}{r \sin \theta} \rightarrow 1\)

Since there is no acceleration in the vertical direction, the vertical component is equal and opposite to the weight of the mass so, the vertical component of tension is

\(T \cos \theta=m g\)

We can rearrange the above equation as

\(T=\frac{m g}{\cos \theta} \rightarrow 2\)

Equating 1 and 2

\(\frac{m v^{2}}{r \sin \theta}=\frac{m g}{\cos \theta}\)

\(v^{2}=\frac{g r \sin \theta}{\cos \theta}\)

\(v^{2}=g r \tan \theta \rightarrow 3\)

Given that, The radius of the circular path, \(r=0.4 m\)

The conical pendulum makes an angle \(\theta=45^{\circ}\)

The acceleration due to gravity, \(g=9.8 m / s\)

Substitute these given values in the equation 3

\(v^{2}=g r \tan \theta\)

\(v^{2}=9.8 \times 0.4 \times \tan 45^{\circ}\)

\(v^{2}=9.8 \times 0.4 \times 1\)

\(v^{2}=3.9=4 m / s\)

\(v=\sqrt{4} m / s\) \(=2 m / s\)

Given,

\(KE _{1}=50 J\)

\(KE _{2}=150 J\)

\(t =10 s\)

As we know,

The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy.

\( W =\Delta KE\)

Where \(W=\) work done and \(\Delta K E=\) change in kinetic energy

By the work-energy theorem,

\( W=\Delta KE \)

\(\therefore W=K E_{2}-K E_{1} \)

\(\Rightarrow W=150-50 \)

\(\Rightarrow W=100 J\)

So, the power of the body \( (P)=\frac{W}{t}\)

Where \(W=\) work and \(t=\) time

\(\therefore P=\frac{100}{10} \)

\(\Rightarrow P=10 W\)

The transfer of electrons from one body to the other results is development of charges. So, no. of electrons given by one body \(=\) No. of electrons obtained by the other.

So, mass of negatively charged body slightly increases while mass of positively charged body slightly increases, whereas the total mass of the system remains the same.

For a diamagnetic material, \(\mu_{\mathrm{r}}\) and \(\epsilon_{\mathrm{r}}\) should have following bounds \(0<\mu_{\mathrm{r}}<1\) and for any material \(\epsilon_{\mathrm{r}}>1\).Diamagnetic materials create an induced magnetic field in a direction opposite to an external applied magnetic field and are repelled by applied magnetic field.Magnetic permeability of the diamagnetic materials is little less than unity.

Accuracy of measurement is determined by the absolute error. Absolute error is the is the difference between the actual and measured value. It is the maximum possible error that needs to be eliminate to get an accurate measurement.