Physics Test - 31

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Physics Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The surface tension of the soap solution is 0.03 N/m. The work done in blowing to form a soap bubble of surface area 40 cm², (in J), is

A
1. 2 x 10^-4

B
2. 4 x 10^-4

C
12 x 10^-4

D
24 x 10^-4

Solution

In case of soap bubble \(\mathrm{W}=\mathrm{T} \times 2 \times \Delta \mathrm{A}=0.03 \times 2 \times 40 \times 10^{-4}=2.4 \times 10^{-4} \mathrm{J}\)
Question 2
1 / -0

Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are β_G, β_R and β_B, respectively. Then,

A
β_G>β_B>β_R

B
β_B>β_G>β_R

C
β_R>β_B>β_G

D
β_R>β_G>β_B

Solution

We know that fringe width is given by
\begin{array}{l}
\because \lambda_{\mathrm{R}}>\lambda_{\mathrm{G}}>\lambda_{\mathrm{B}} \\
\therefore \beta_{\mathrm{R}}>\beta_{\mathrm{G}}>\beta_{\mathrm{B}}
\end{array}
Question 3
1 / -0

The circuit shown here is used to compare the emf'semf′s of two cells E1 and E2(E1>E2). The null point is at C when the galvanometer is connected to E₁. When the galvanometer is connected to E₂, the null point will be -

A
to the left of C

B
to the right of C

C
at C itself

D
no where on AB

Solution

\(\mathbf{E}_{1}=\phi \times 1_{1} \ldots\)
(i) Given \(\mathrm{E}_{2}<\mathrm{E}_{1}\)
\(\mathrm{E}_{2}=\phi \times 1_{2} \ldots\)
(ii) Therefore \(l_{2}Hence balance point will shift towards left of \(C\).
KEY CONCEPTS
Meter Bridge
- Meter bridge is based on the principle of Wheatstone
- bridge and it is used to find the resistance of an unknown conductor or to compare two unknown
- Resistance
- Figure below shows a schematic diagram of a meter bridge

In above figure \(A C\) is a \(1 \mathrm{m}\) long wire made of manganin or constantan having uniform area of cross-section
This wire is stretched along a scale one a wooden base
- Ends A and C of the wire are screwed to two L shaped copper strips as shown in figure
- A resistance box \(\mathrm{R}\) and an unknown resistance \(\mathrm{S}\) are connected as shown in figure
One terminal of galvanometer is connected to point \(\mathrm{D}\) and another terminal is joined to a jockey that can be slided on a bridge wire
- When we adjust the suitable resistance of value \(R\) in the resistance box and slide this jockey along the wire then a balance point is obtained sat at point \(B\)
since the circuit now is the same as that of Wheatstone bridge . so from the condition of balanced Wheatstone bridge we have
\(-\frac{p}{Q}=\frac{R}{s}\)
- Here resistance P equals
\(-P=\frac{\rho l_{1}}{A}\)
And \(Q=\frac{\rho_{2}^{2}}{A}-\) Where \(\rho\) is the resistivity of the material of the wire and \(A\) is the area of cross section of wire- Now \(\frac{P}{Q}=\left(\frac{\rho l_{1}}{A}\right)\left(\frac{A}{\rho l_{2}}\right)=\frac{l_{1}}{l_{2}}\)
Examples on Meter Bridge

I. Find the resistance per unit length of the connecting wire \(A C\).
in. Suppose at point \(B\) the deflection in galvanometer is recorded zero, at that point the wheatstone bridge is balanced.
iii. We can see that \(\frac{R}{R_{a}}=\frac{S}{R_{l a}},\) using this carry out further calculations.
Example
In meter bridge the balancing length from left and when standard resistance of \(1 \Omega\) is in right gas is found to be \(20 \mathrm{cm}\). The value of unknown resistance is:

In meter bridge the balancing length from left and when standard resistance of \(1 \Omega\) is in right gas is found to be \(20 \mathrm{cm}\). The value of unknown resistance is:
Solution The condition of wheatstone bridge gives \(\frac{x}{R}=\frac{20 r}{80 r}\)
r- resistance of wire per \(\mathrm{cm}\)
X- unknown resistance
\(\therefore \mathrm{X}=\frac{1}{4} \times R=\frac{1}{4} \times 1=0.25 \Omega\)
Question 4
1 / -0

A strip of copper and another of germanium are cooled from room temperature to 80 K. The resistance of:

A
Each of these decreases

B
Copper strip increases and that of germanium decreases

C
Copper strip decreases and that of germanium increases

D
Each of these increases

Solution

Copper is conductor and germanium is semiconductor. When cooled, the resistance of copper strip decreases and that of germanium increases.
Question 5
1 / -0

Solid metallic cuboid of homogeneous material having geometrical dimension as shown in figure will have ratio of resistances for the sense of current along X, Y and Z directions respectively-

A
a : b : c

B
a : c : b

C
a² : b² : c²

D
a² : c² : b²

Solution

\(\mathrm{R}_{\mathrm{X}}=\frac{\rho \mathrm{a}}{\mathrm{bc}}\)
\(\mathrm{R}_{\mathrm{Y}}=\frac{\rho \mathrm{c}}{\mathrm{ab}}\)
\(\mathrm{R}_{\mathrm{Z}}=\frac{\rho \mathrm{b}}{\mathrm{ac}}\)
\(\mathrm{R}_{\mathrm{X}}: \mathrm{R}_{\mathrm{Y}}: \mathrm{R}_{\mathrm{Z}}:: \frac{\mathrm{a}}{\mathrm{bc}}: \frac{\mathrm{c}}{\mathrm{ab}}: \frac{\mathrm{b}}{\mathrm{acc}}\)
\(\mathrm{a}^{2}: \mathrm{c}^{2}: \mathrm{b}^{2}\) (multiply each ratio by abc )
KEY CONCEPTS

Resistance
I. The property of substance by virtue of which it opposes the flow of current through it, is known as the resistance, \(\mathbf{R}=\frac{\rho \mathbf{l}}{\mathbf{A}}\)
ii. Unit and dimension:lt's S.I. unit is Volt/Amp or ohm(W). Also 1 ohm \(=10^{9}\) emu of resistance. Its dimension is \(\left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-3} \mathrm{A}^{-2}\right]\)
iii. Dependence of resistance: Resistance of a conductor depends upon the following factors.
(i) Length of the conductor: Resistance of a conductor is directly proportional to it's length i.e., \(\mathrm{R} \propto l\) and inversely proportional to it's area of cross-section i.e. \(R \propto \frac{1}{A}\)
(ii) temperature: For a conductor Resistance \(\alpha\) Temperature
Solved Examples:
Question 1:
Calculate the resistance offered by the body if 6 A of current is flowing having potential difference of 2 V?
Solution: Given: Current \(1=6 \mathrm{A},\) Potential difference \(\mathrm{V}=2 \mathrm{V}\) The resistance is given by \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=0.33 \Omega\)
Question 2:
Calculate the resistance if \(8 \Omega\) and \(9 \Omega\) are connected in parallel.
Solution: Given: resistance \(R_{1}=8 \Omega, \mathrm{R}_{2}=9 \Omega\) the parallel resistance is given by
\(R_{P}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{(9 \times 8)}{17}=4.235 \Omega\)
Question 6
1 / -0

A particle at the end of a spring executes simple harmonic motion with a period t₁, while the corresponding period for another spring is t₂. If the period of oscillation with the two springs in series is T, then

A

B

C

D

Solution
Question 7
1 / -0

A merry-go-round is at rest, pivoted about a frictionless axis. A child of mass m runs along the path tangential to the rim with speed v and jumps on to merry-go-round. If R be the radius of merry-go-round and I is the moment of inertia, then the angular velocity of the merry-go-round and the child is

A

B

C

D

Solution
Question 8
1 / -0

A bar magnet has a magnetic moment of 10 A m². If its magnetic length is 5 cm, Its pole strength is

A
200 A m

B
100 A m

C
300 A m

D
50 A m

Solution

\(m=\frac{M}{l}=\frac{10}{0.05}=200 \mathrm{A} \mathrm{m}\)
KEY CONCEPTS
Magnetic field due to broad side-on position Magnetic field on the equatorial position (Braod side on position) isWhen point lies on equatorial line of magnetic dipole, it is called as broadside position or tangent B position. On equatorial position:
\(B_{e}=\frac{\mu_{0}}{4 \pi} \frac{M}{\left(r^{2}+I^{2}\right)^{\frac{2}{2}}} ;\) if \(l \ll r,\) then \(B_{e}=\frac{\mu_{0}}{4 \pi} \frac{M}{r^{2}}\)
Current carrying loop as a magnetic dipole A loop as a magnet
The pattern of the magnetic field is comparable with the magnetic field produced by a bar magnet.

The side "I' (the side from the \(\vec{B}\) emerges out) of the loop acts as 'NORTH POLE' and side I (the side in which the \(\vec{B}\) entries) acts as the 'SOUTH POLE'. It can be verified by studying force on one loop due to a magnet or a loop.

\(B_{a d s} \frac{\mu_{0} N I R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}} \cong \frac{\mu_{0} N I R^{2}}{2 x^{3}}\) for \(x \gg R\)
\(=2\left(\frac{\mu_{0}}{4 \pi}\right)\left(\frac{I N \pi R^{2}}{x^{3}}\right)\)
it is similar to \(\mathrm{B}_{\text {axis }}\) due to magnet \(=2\left(\frac{\mu_{0}}{4 \pi}\right) \frac{\mathrm{m}}{x^{2}}\) Magnetic dipole moment of the loop
\(M=I N \pi R^{2}\)
\(\mathrm{M}=\) INA for any other shaped loop Unit of \(\mathrm{M}\) is Amp. \(\mathrm{m}^{2}\) Unit of \(\mathrm{m}\) (pole strength) \(=\) Amp. \(\mathrm{m}\)
\(\{\because \text { in magnet } M=m \mathrm{l}\}\)
\(\vec{M}=I N \vec{A}\)
\(\vec{A}=\) unit normal vector for the loop. To be determined by right hand rule which is also used to determine direction \(\vec{B}\) on the axis. It is also from 'S' side to "N' side of the loop.
Question 9
1 / -0

The time period of a satellite of earth is 5 hour. If the seperation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

A
10 hour

B
80 hour

C
40 hour

D
20 hour

Solution

According to Kepler's law \(T^{2} \propto r^{3}\)

\(\therefore\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{r_{1}}{r_{2}}\right)^{3}=\left(\frac{1}{4}\right)^{3}=\frac{1}{64}\) or \(\frac{T_{1}}{T_{2}}=\frac{1}{8}\)

or \(T_{2}=8 T_{1}=8 \times 5=40\) hour

KEY CONCEPTS

Examples on Third law The mean distance from the Earth to the Sun is \(1.496 \times 10^{8} \mathrm{km}\) and the period of its motion about the Sun is 1.00 year. The period of Jupiter's motion about the Sun is observed to be 11.86 years. Determine the mean distance from the Sun to the Jupiter.
Question 10
1 / -0

According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope

A
depends on the nature of the metal used

B
depends on the intensity of the radiation

C
depends both on the intensity of the radiation and the metal used

D
is the same for all metals and is independent of the intensity of the radiation

Solution

According to Einstein's equation,
Kinetic energy \(=h f-\phi\) where kinetic energy and \(f\) (frequency) are variables, compare it with equation, \(y=m x+c\)

slope of line \(=h\) (h is Planck's constant.)
Hence the slope is same for all metals and independent of the intensity of radiation.