Physics Test - 11

For the body to move with uniform speed, resultant force on the body should be zero.

Therefore, the force applied on the body is equal to the weight force of the body along the plane.

\(F=m g \sin \theta\)

\(100=m(10) \sin 30^{\circ}(5)\)

\(m=\frac{100}{(10)(0.5)(5)}=4 {~kg}\)

Now the force required to move the block up the plane with an acceleration of \(4 {~ms}^{-2}\) is:

\({F}-{mg} \sin \theta={ma}\)

\(F=m(a+g \sin \theta)=4(4+10(0.5))=36 {~N}\)

Work done \(=36 \times 5=180 {~J}\)

Given,

\(A=2.5 \mathrm{~ms}^{-1} \pm 0.5 \mathrm{~ms}^{-1}\)

\( B=0.10 \mathrm{~s} \pm 0.01 \mathrm{~s}\)

\(Z=A B=(2.5)(0.10)=0.25 \mathrm{~m}\)

Then we can write,

\(\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}\)

\(\Delta Z=Z\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right)\)

\(=0.25\left(\frac{0.5}{2.5}+\frac{0.01}{0.10}\right)\)

\(=0.25(0.2+0.1)=0.075\)

\(\Delta Z=0.075=0.08 \mathrm{~m}\)

Thus, measured value of \(A B\), i.e., \(=Z \pm \Delta Z=(0.25 \pm 0.08) \mathrm{m}\)

Given,

\({V}=3 {~V}\)

\({R}_{1}=10 \Omega\)

\({R}_{2}=20 \Omega\)

Let, \(\mathrm{D}_{1}\) is the diode connected in series with resistance \(\mathrm{R}_{1}\) and \(\mathrm{D}_{2}\) is the diode connected in series with resistance \(R_{2}\)

Here, both the diodes are ideal hence, when the positive terminal of the battery is connected to \(B\) diode \(\mathrm{D}_{1}\) becomes reverse biased and offers infinite resistance to the current. Also, diode \(\mathrm{D}_{2}\) becomes forward biased and offers zero resistance to the current.

Therefore current will flow only through the resistance \(R_{2}\).

\(I=\frac{V}{R_{2}}\)

\(I=\frac{3}{20}\)

\(I=0.15 {~A}\)

Bernoulli's Principle: It states the total mechanical energy of the moving fluid comprising the energy associated with the fluid pressure (P), the gravitational potential energy of elevation ρgh, and the kinetic energy of the fluid motion \(\frac{1}{2} \rho v^{2}\), remains constant. i.e.,

\(P+\frac{1}{2} \rho v^{2}+\rho g h=\) const.

Where p is the pressure exerted by the fluid, v is the velocity of the fluid, ρ is the density of the fluid, h is the height of the container.

Given,

The mass of the big particle is \(4 {~m}\) and velocity of two fragments are mutually perpendicular to each other. If \(p_{1}\) and \(p_{2}\) are the momentum of the fragments then momentum of third fragment is:

\({p}_{3}=\sqrt{{p}_{1}^{2}+{p}_{2}^{2}}=\sqrt{({mv})^{2}+({mv})^{2}}\) \(=\sqrt{2} {mv}\)

Velocity of third fragment \(=\frac{{v}}{\sqrt{2}}\)

Total energy released in the explosion:

\(=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2} \times 2 m\left(\frac{v}{\sqrt{2}}\right)^{2}\)

\(=\frac{3}{2} m v^{2}\)

Thus, the total energy released in the explosion is \(\frac{3}{2} {mv}^{2}\)

According to Charles' law if the pressure of the gas is constant,

\(V \propto T\)

Charles' law: Charles' law states that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature if the pressure remains constant.

It is the largest distance between a source and a destination up to which the signal is received with sufficient strength.

The loss of strength of a signal while propagating through a medium is known as attenuation.

Bandwidth refers to the frequency range over which a piece of equipment operates or the portion of the spectrum occupied by the signal.

We know that the largest distance between a source and a destination up to which the signal is received with sufficient strength is called the range of the communication system.

Sensitivity of a moving coil galvanometer is defined as the current \(i\) in microampere required to consume \(1\) millideflection.

∴ Sensitivity

\(k= \frac{θ}{i}\)

\(= \frac{NBA}{C}\)​

Where all the symbols have their usual meaning and \(C\) is the torque per unit twist

Thus the sensitivity of the galvanometer can be increased either by increasing the number of turns \((N)\), using strong magnet or increasing the area of the coil .

Given,

Magnification of compound microscope \((m)=60\)

Magnification of objective lens \(\left(m_{o}\right)=10\)

Image is formed at near point \(({V}{D})=25 {~cm}\).

Linear magnification due to eyepiece:

\(m_{e}=1+\frac{D}{f_{e}}=1+\left(\frac{25}{f_{e}}\right)\)

Where \({f}_{{e}}\) is the focal length of the eyepiece...(i)

Total magnification is given by:

\({m}={m}_{{o}} \times {m}_{{e}}\)....(ii)

Substituting (i) in (ii)

\(60=m_{o}\left(1+\frac{D}{f_{e}}\right)=10\left(1+\left(\frac{25}{f_{e}}\right)\right) \)

\(50=\left(\frac{250}{f_{e}}\right) \)

\({f}_{{e}}=5 {~cm} \)

Beta rays are not electromagnetic waves because they are charged particles and are capable of getting deflected by the magnetic field. These rays are not pure energy as a photon