Physics Test

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Physics Test - 11

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Weekly Quiz Competition

Question 1
4 / -1

A body moves with a speed of \(10\) m/s in the curved path of \(25\) m radius of curvature. If the tangential acceleration is \(3\) m/s\(^2\), then total acceleration for the body will be:

A
\(3.3\) m/s\(^2\)

B
\(4\) m/s\(^2\)

C
\(5\) m/s\(^2\)

D
\(1\) m/s\(^2\)

Solution

Given:
\(v=10\)m/s, \(r=25\) m and\({a}_{{t}}=3\)m/s\(^2\)
Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,
\(a=\sqrt{a_{c}^{2}+a_{t}^{2}}\)
Centripetal Acceleration \(\left({a}_{{c}}\right)\):
\(\therefore a_{c}=\frac{v^{2}}{r}\)
\(\Rightarrow a_{c}=\frac{(10)^{2}}{25}=\frac{100}{25}\)m/s\(^2\)
Thus, net acceleration
\(a=\sqrt{a_{t}^{2}+a_{c}^{2}}\)
\(=\sqrt{4^{2}+3^{2}}=5\)m/s²
Question 2
4 / -1

In the series LCR circuit, the power dissipation is through:

A
R

B
L

C
C

D
Both L and C

Solution

In the series LCR circuit, the power dissipation is through R.

The capacitor and inductor are the storage devices that store the energy in it. The inductor and capacitor can’t dissipate energy. As the resistance opposes the flow of electric current. So the resistance of any circuit dissipates the power.
Question 3
4 / -1

A solution containing active cobalt \({ }_{27}^{60} \mathrm{Co}\) having activity of \(0.8 \mu \mathrm{Ci}\) and decay constant \(\lambda\) is injected in an animal's body. If \(1 \mathrm{~cm}^{3}\) of blood is drawn from the animal's body after 10 hrs of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body ? \(\left(1 \mathrm{Ci}=3.7 \times 10^{10}\right.\) decays per second and at \(t=10\) hrs \(\left.e^{-\lambda t}=0.84\right)\)

A
6 liters

B
7 liters

C
4 liters

D
5 liters

Solution

The activity equation can be written as
\(-\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}_{\circ} \mathrm{e}^{-\lambda \mathrm{t}}\)
Given that
\(\lambda \mathrm{N}_{0}=0.8 \mu \mathrm{C}_{\mathrm{i}}\)
Putting the values,
\(\lambda \mathrm{N}_{0}=2.96 \times 10^{4}\)
Let the volume of the blood flowing be \(\mathrm{V}\),
the activity would reduce by a factor of \(\frac{10^{-3}}{\mathrm{~V}}\)
Hence,
\(\frac{\lambda \mathrm{N}_{0} 10^{-3}}{\mathrm{~V}} \mathrm{e}^{-\lambda \mathrm{t}}=\frac{300}{ 60}\) (Both R.H.S. and L.H.S. are decay)
Putting the values of \(\mathrm{e}^{-\lambda \mathrm{t}}\) and \(\lambda \mathrm{N}_{0}\) we get
\(\mathrm{V}=5\)liters
Question 4
4 / -1

As shown in the figure, two infinitely long, identical wires are bent by \(90^{\circ}\) and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If \({OP}={QQ}=4 {~cm},\) and the magnitude of the magnetic field at \({O}\) is \(10^{-4} {~T},\) and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at \({O}\) will be: \(\left(\mu_{0}=4 \pi \times 10^{-7} {NA}^{-2}\right)\)

A
20 A, perpendicular into the page

B
40 A, perpendicular out of the page

C
20 A, perpendicular out of the page

D
40 A, perpendicular into the page

Solution

Magnetic field at ' \(O^{\prime}\) will be done to 'PS' and 'QN' only
i.e \(\mathrm{B}_{0}=\mathrm{B}_{\mathrm{PS}}+\mathrm{B}_{\mathrm{QN}} \rightarrow\) Both inwards
Let current in each wire \(=i\)
\(\therefore B_{0}=\frac{\mu_{0} i}{4 \pi d}+\frac{\mu_{0} i}{4 \pi d}\)
\(\therefore i=20 \mathrm{~A}\)
Question 5
4 / -1

Power is define as________.

A
it is the push or pull of an object

B
the capacity to do work

C
the rate of doing work

D
None of the above

Solution

Power (P): The rate of work done is called power.
The SI unit of power is the watt (W).
Question 6
4 / -1

The value of universal gravitational constant \(\mathrm{G}\) is ________.

A
\(6.67 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}}\)

B
\(6.67 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}^{2}}\)

C
\(66.7 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}^{2}}\)

D
\(66.7 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}}\)

Solution

Gravitational force between two objects of mass \(m_{1}\) and \(m_{2}\) separated by a distance \(r, F=\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
\[\Rightarrow \mathrm{G}=\frac{F \mathrm{r}^{2}}{\mathrm{~m}_{1} \mathrm{~m}_{2}}\]
Value of universal gravitational constant \(\mathrm{G}\) is \(6.67 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}^{2}}\).
Question 7
4 / -1

In the relation \(p=\frac{\alpha}{\beta} \mathrm{e}^{-\frac{\alpha z}{k \theta}}, p\) is the pressure, \(z\) the distance, \(k\) is Boltzmann constant and \(\theta\) is the temperature, the dimensional formula of \(\beta\) will be:

A
\(\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\)

B
\(\left[\mathrm{ML}^{2} \mathrm{~T}\right]\)

C
\(\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]\)

D
\(\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\)

Solution

The given relation,
\(p=\frac{\alpha}{\beta} \mathrm{e}^{-\frac{\alpha z}{k \theta}}\)
\(pe^0=\frac{\alpha}{\beta} \mathrm{e}^{-\frac{\alpha z}{k \theta}}\)
By comparing both sides, we get
\(\frac{\alpha z}{k \theta} = 0\)
\(\Rightarrow \alpha=\frac{k \theta}{z}\)
\(\Rightarrow [\alpha]=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1} \times \mathrm{K}\right]}{[\mathrm{L}]}\)
\(=\left[\mathrm{MLT}^{-2}\right]\)
And \( p=\frac{\alpha}{\beta}\)
\(\Rightarrow [\beta]=\left[\frac{\alpha}{p}\right]\)
\(=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}\)
\(=\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\)
Question 8
4 / -1

Energy stored in a parallel plate capacitor is:

A
\(\frac{3 \mathrm{q}^{2}}{2 \mathrm{C}}\)

B
\(\frac{\mathrm{q}^{2}}{\mathrm{C}}\)

C
\(\frac{\mathrm{q}^{2}}{2 \mathrm{C}}\)

D
\(\frac{\mathrm{q}^{2}}{3\mathrm{C}}\)

Solution

The capacitor is a charge storage device work has to be done to store the charges in a capacitor. This work done is stored as electrostatic potential energy in the capacitor.
Let \(\mathrm{q}\) be the charge and \(\mathrm{V}\) be the potential difference between the plates of the capacitor. If dq is the additional charge given to the plate,
Work done is,
\(\mathrm{dw}=\mathrm{V}\) dq. \(\mathrm{dw}\)
\(=\frac{\mathrm{q}}{\mathrm{C}} \mathrm{dq} \quad\left(\therefore \mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}\right)\)
Total work done to charge a capacitor is,
\(\mathrm{w}=\int \mathrm{d} \mathrm{w}\)
\(=\int_{0}^{\mathrm{q}} \frac{\mathrm{q}}{\mathrm{C}} \mathrm{dq}\)
\(=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}\)
This work done is stored as electrostatic potential energy (U) in the capacitor.
\(\mathrm{U}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}=\frac{1}{2} \mathrm{CV}^{2} \quad(\therefore \mathrm{q}=\mathrm{CV})\)
\(\mathrm{U}=\frac{\mathrm{q}^{2}}{\mathrm{2 C}}\)
Question 9
4 / -1

Two bodies of masses 4 kg and 5 kg are acted upon by the same force if acceleration of lighter body is 3m/s²

A
1.4 m/sec²

B
2.4 m/sec²

C
3.4 m/sec²

D
4.4 m/sec²

Solution

For lighter body
\({F}={M} \times {a}\)
\({F}=4 \times {3}=12 {N}\)
For heavier body, since the force is same
\(12=5 \times {a}\)
\({a}=\frac{12}{5}\)
\({a}=2.4 \) m/sec²
Question 10
4 / -1

The period of oscillation of a mass \(M\) suspended from a spring of negligible mass is \(T\). If along with it another mass \(M\) is also suspended figure (b), the period of oscillation will now be:

A
\(T\)

B
\(\frac{T}{\sqrt{2}}\)

C
\(2T\)

D
\(\sqrt{2} T\)

Solution

In figure (a) a block of mass \(M\) is suspended with a massless spring constant \(k\).

Thus the time period is given as:

Time period \(T=2 \pi \sqrt{\frac{M}{K}} \ldots \ldots . .\) \((i)\)

In figure (b) two blocks of mass \(M\) are also suspended

Therefore time period for the second spring:

\(\begin{aligned} T^{\prime} &=2 \pi \sqrt{\frac{M+M}{K}} \\ &=2 \pi \sqrt{\frac{2 M}{K}} \\=& \sqrt{2} T \end{aligned}\)

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Physics Test - 11

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Physics Test - 11

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Question 1

\(3.3\) m/s\(^2\)

\(4\) m/s\(^2\)

\(5\) m/s\(^2\)

\(1\) m/s\(^2\)

Solution

Question 2

R

L

C

Both L and C

Solution

Question 3

6 liters

7 liters

4 liters

5 liters

Solution

Question 4

20 A, perpendicular into the page

40 A, perpendicular out of the page

20 A, perpendicular out of the page

40 A, perpendicular into the page

Solution

Question 5

it is the push or pull of an object

the capacity to do work

the rate of doing work

None of the above

Solution

Question 6

\(6.67 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}}\)

\(6.67 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}^{2}}\)

\(66.7 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}^{2}}\)

\(66.7 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}}\)

Solution

Question 7

\(\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\)

\(\left[\mathrm{ML}^{2} \mathrm{~T}\right]\)

\(\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]\)

\(\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\)

Solution

Question 8

\(\frac{3 \mathrm{q}^{2}}{2 \mathrm{C}}\)

\(\frac{\mathrm{q}^{2}}{\mathrm{C}}\)

\(\frac{\mathrm{q}^{2}}{2 \mathrm{C}}\)

\(\frac{\mathrm{q}^{2}}{3\mathrm{C}}\)

Solution

Question 9

1.4 m/sec2

2.4 m/sec2

3.4 m/sec2

4.4 m/sec2

Solution

Question 10

\(T\)

\(\frac{T}{\sqrt{2}}\)

\(2T\)

\(\sqrt{2} T\)

Solution

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1.4 m/sec²

2.4 m/sec²

3.4 m/sec²

4.4 m/sec²