Physics Test - 24

Given,

\(V=(100 \pm 5) \mathrm{V}\)

\(I=(10 \pm 0.2) \mathrm{A}\)

The resistance is given by,

\(R=\frac{V}{I}\)

\(\frac{\Delta R}{R}=\left(\frac{\Delta V}{V}+\frac{\Delta I}{I}\right)\)

For calculating percentage error,

\((\frac{\Delta R}{R} \times 100) \%= (\frac{\Delta V}{V} \times 100+\frac{\Delta I}{I} \times 100) \%\)

\(= (\frac{5}{100} \times 100+\frac{0.2I}{10} \times 100) \%\)

\(= 7 \%\)

Simple harmonic motion (S.H.M) is a type of periodic oscillation where the restoring force is directly proportional to the displacement.

The motion of a simple pendulum is a good example of simple harmonic motion in which bob is oscillating freely along with the lowest and highest position under the influence of gravity.

From the above explanation, we can see that, for a simple harmonic motion restoring force is directly proportional to its displacement from its mean position

i.e., f ∝ -x

⇒ f = -kx

Here, f is restoring force, x is displacement and k is proportionality constant.

Also in the above expression, a negative sign represents that restoring for and displacement is in the opposite directions.

Air has the least thermal conductivity.

The thermal conductivity of a material is a measure of the ability of the material to conduct heat.A high value for thermal conductivity indicates that the material is a good heat conductor, and a low value indicates that the material is a poor heat conductor or insulator. Brass and mercury are metals so they have a high value of conductivity, whereas air has the least.

i.e., \(K\)_air \(< K\)_water \(< K\)_mercury \(< K\)_brass

The amplitude of torque is as follows,

\(\Rightarrow[T]=\left[M^{1} L^{2} T^{-2}\right]\)

The dimension of the work is as follows,

\(\Rightarrow[T]=\left[M^{1} L^{2} T^{-2}\right]\)

The amplitude of the momentum is as follows,

\(\Rightarrow[\mathrm{T}]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]\)

From the above explanation, it is clear that torque and work have same dimension but momentum has a different dimension.

Hooke's law states that within the elastic limit stress is directly proportional to strain.

Hooke's law states that " Within the elastic limit stress is directly proportional to strain" The stress of an elastic material is the restoring force acting per unit area of an object. The strain is the ratio of change in dimension to the original dimension. The ratio of stress to strain is known as an elastic modulus.

Given: mass of glass \(m_{1}=100\) g, initial temperature \(t_{1}=350^{\circ}\) C, specific heat capacity of the glass \(c _{1}=0.8\) J/g\(^{\circ}\) C and mass of water \(m _{2}=1\) kg \(=1000\) g, final temperature \(t _{2}=23.6^{\circ}\) C, specific heat capacity of water \(C _{2}=4.0\) J/g\(^{\circ}\) C

Let the final temperature of the system will be \(x ^{\circ}\) C.

Heat lost by glass = Heat gained by water

\(m_{1} c_{1}\left(t_{1}-x\right)=m_{2} c_{2}\left(x-t_{2}\right)\)

\(100 \times 0.8 \times(350-x)=1000 \times 4.0(x-23.6)\)

\(350-x=50(x-23.6)\)

\(51 x=1530\)

\(\Rightarrow x=30^{\circ}\) C

A solenoid is an instrument that consists of copper coiling over a cylinder designed to create a strong magnetic field inside the coil because of the flow of current through the coil. By wrapping the same wire many times around a cylinder, the magnetic field due to the flow of current can become quite strong. So, we can say that the strength of the magnetic field will change as a current through the coil or the number of turn’s changes. Magnetic field strength is independent of the diameter of the cylinder of a solenoid. The strength of the magnetic field in a solenoid will be directly proportional to the number of turns and amount of current flowing through a wire and will be inversely proportional to its length.

Thus it is given by \(B = \frac{{{μ _0}NI}}{l}\)

Where, \(N =\) number of turns and \(I =\) current, \(l =\) length of the solenoid

Given:

Length of 1st solenoid = L, Length of 2nd solenoid = 2L, number of turns of 1st solenoid = N and number of turns of 2nd solenoid = 4L

• Magnetic field due to the solenoid is given by

\(\Rightarrow B = \frac{{{μ _0}NI}}{l}\)

• As μo and current (I) is constant, therefore

\(\Rightarrow \;B \propto \frac{N}{L}\;\)

\( \Rightarrow \;\frac{{{B_1}}}{{{B_2}}}\; = \;\frac{{{N_1}}}{{{N_2}}} \times \frac{{{L_2}}}{{{L_1}}}\; = \;\frac{N}{{4N}} \times \frac{{2l}}{L}\; = \;\frac{1}{2}\)

Given that, \(r=50 cm , R=100 cm\)

Mass supported on four columns, \(M=50 \times 10^3 kg\)

Mass supported on each column, \(m=\frac{M}{4}\)

\(\Rightarrow m=\frac{50 \times 10^3}{4}=12.5 \times 10^3 kg\)

Now, weight, \(w=m g=12.5 \times 9.8 \times 10^3 N =1225 \times 10^5 N\)

Area of cross-section of each column

\(A =\pi\left( R ^2-r^2\right) \)

\( =3.14\left\{(100)^2-(50)^2\right\} \times 10^{-4} m ^2=2.35 m ^2\)

Young's modulus, \(Y =2.0 \times 10^{11} Pa\)

By using Hooke's law,

Stress \(=Y \times\) Strain

\(\therefore\) Compressive strain \(=\frac{\text { Stress }}{ Y }=\frac{ W }{ AY }\)

Substituting the values, we get

Compressive strain \(=\frac{1.225 \times 10^5}{2.35 \times 2.0 \times 10^{11}}=2.60 \times 10^{-7}\)