Physics Test

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Physics Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process ABCA will be :

A
25%

B
12.5%

C
50%

D
$\frac{100}{13} %$

Solution

$\mathrm{W}=\frac{1}{2} \mathrm{P}_{0} \mathrm{V}_{0}=\frac{1}{2} \mathrm{RT}_{0}$

Heat absorbed $=\mathrm{Q}_{\mathrm{AB}}+\mathrm{Q}_{\mathrm{BC}}=\mathrm{C}_{\mathrm{V}} \mathrm{T}_{0}+\mathrm{C}_{\mathrm{p}} 2 \mathrm{T}_{0}=\frac{13}{2} \mathrm{RT}_{0}$
Efficiency $\left.=\frac{\frac{1}{2} \mathrm{P}_{0} \mathrm{V}_{0}}{\frac{13}{2} \mathrm{P}_{0} \mathrm{V}_{0}} \cdot: \frac{13}{2} \mathrm{P}_{0} \mathrm{V}_{0}=\frac{13}{2} \mathrm{RT}_{0}\right]$
$=\frac{1}{13} \times 100=7.7 \%$
Hence, the correct option is D.
Question 2
1 / -0

Two identical conducting spheres each having radius r are placed at large distance. lnitially charge on one sphere is q, while charge on another sphere is zero when they are connected by conducting wire as shown in figure then find total heat produced when switch S is closed :

A
$\frac{k q^{2}}{2 r}$

B
$\frac{3 \mathrm{Kq}^{2}}{2 \mathrm{r}}$

C
$\frac{\mathrm{Kq}^{2}}{4 \mathrm{r}}$

D
zero

Solution

When sphere connected by wire, charge will flow from one sphere to another until potential of both sphere become same.Or,

As both sphere is identical in dimension and material, charge will divide equally

Now, Initial potential energy of system

$\mathrm{U}_{\mathrm{i}}=\frac{\mathrm{K}\left(\mathrm{Q}^{2}\right)}{2 \mathrm{r}}($ Self energy of spherical conductor $)$

Final potential energy of system $\mathrm{U}_{\mathrm{f}}=\frac{\mathrm{K}(\mathrm{Q} / 2)^{2}}{2 \mathrm{r}}+\frac{\mathrm{K}(\mathrm{Q} / 2)^{2}}{2 \mathrm{r}}$

$=\frac{\mathrm{KQ}^{2}}{4 \mathrm{r}}$
Heat generation in conduction $=U_{i}-u_{f}$ $=\frac{\mathrm{KQ}^{2}}{2 \mathrm{r}}-\frac{\mathrm{KQ}^{2}}{4 \mathrm{r}}$
$=\frac{\mathrm{KQ}^{2}}{4 \mathrm{r}}$
Hence, the correct option is C.
Question 3
1 / -0

A soap bubble (surface tension = T) is charged to a maximum surface density of charge = σ, when it is just going to burst. Its radius R is given by:

A
$\mathrm{R}=\frac{\sigma^{2}}{8 \varepsilon_{0} \mathrm{T}}$

B
$\mathrm{R}=8 \varepsilon_{0} \frac{\mathrm{T}}{\sigma^{2}}$

C
$\mathrm{R}=\frac{\sigma}{\sqrt{8 \varepsilon_{0} \mathrm{T}}}$

D
$R=\frac{\sqrt{8 \varepsilon_{0} T}}{\sigma}$

Solution

Pressure due to surface tension $=\frac{4 \mathrm{T}}{\mathrm{R}}$ (soap bubble) Pressure due to electrostatic force $=\frac{\sigma^{2}}{2 \varepsilon_{0}}$ Just before the bubble burst

Pressure due to surface tension = Pressure due to electrostatic force $\Rightarrow \frac{4 \mathrm{T}}{\mathrm{R}}=\frac{\sigma^{2}}{2 \varepsilon_{0}}$
$\Rightarrow R=\frac{8 \mathrm{T} \varepsilon_{0}}{\sigma^{2}}$
Hence, the correct option is B.
Question 4
1 / -0

Magnetic moments of two identical magnets are M and 2M respectively. Both are combined in such a way that their similar poles are same side. The time period in this case is ‘T1’ .If polarity of one of the magnets is reversed its period becomes ‘T2’ then find out ratio of their time periods T1/T2.

A
$\sqrt{3}$

B
$\frac{1}{\sqrt{3}}$

C
1

D
$\frac{1}{3}$

Solution
Question 5
1 / -0

Find the heat generated after the switch S is closed in the following circuit-

A
$\frac{C V^{2}}{2}$

B
$\frac{C V^{2}}{6}$

C
$\frac{2 C V^{2}}{3}$

D
$\frac{C V^{2}}{12}$

Solution
Question 6
1 / -0

An infinite current carrying wire, carrying current I is bent in V shape, lying in x-y plane as shown in figure. Intensity of magnetic field at point P will be (take OP = 2a)

A
$\frac{\mu_{0} I}{4 \pi a}(2+\sqrt{3}) \hat{k}$

B
$\frac{\mu_{0} I}{4 \pi a}(2-\sqrt{3}) \hat{\mathrm{k}}$

C
$\frac{\mu_{0} I}{4 \pi a}(2+\sqrt{3})(-\hat{k})$

D
$\frac{\mu_{0} I}{4 \pi \mathrm{a}}(2-\sqrt{3})(-\hat{\mathrm{k}})$

Solution

$B$ at $P$ due to 1 wire $\overline{\mathrm{B}}=\frac{\mu_{0} i}{4 \pi a}\left[\sin 60^{\circ}+\sin 90^{\circ}\right] \hat{\mathrm{k}}$

$=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}\left(\frac{\sqrt{3}}{2}+1\right) \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{B}}$ due to both wire $=2 \overrightarrow{\mathrm{B}}(\overrightarrow{\mathrm{B}}$ is in same direction) $\bar{B}_{\text {net }}=\frac{\mu_{0} i}{4 \pi a}(2+\sqrt{3}) \hat{k}$
Hence, the correct option is A.
Question 7
1 / -0

A small circular wire loop of radius a is located at the centre of a much larger circular wire loop of radius b as shown above (b> >a). Both loops are coaxial and coplanar. The larger loop carries a time (t) varying current I = I0 cos ωt where I0 and ω are constants. The large loop induces in the small loop an emf that is approximately equal to which of the following.

A
$\frac{\pi \mu_{0} I_{0}}{2} \frac{a^{2}}{b} \omega \cos \omega t$

B
$\frac{\pi \mu_{0} I_{0}}{2} \frac{\mathrm{a}^{2}}{\mathrm{b}} \omega \sin \omega \mathrm{t}$

C
$\frac{\pi \mu_{0} I_{0}}{2} \frac{a}{b^{2}} \omega \sin \omega t$

D
$\frac{\pi !_{0} I_{0}}{2} \frac{a}{b^{2}} \omega \cos \omega t$

Solution

Magnetic flux passing through inner loop due to current in outer loop$\phi=B_{\text {centre }} A_{\text {inner }}(A s, b>r)$

$\phi=\left(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{i}}{\mathrm{b}}\right) \cdot \pi \mathrm{a}^{2}$
$\phi=\left(\frac{\mu_{0} \cdot \pi a^{2}}{2 b}\right) \mathrm{i}_{0} \cos \omega \mathrm{t}$
So, emf developed, $e=\left|\frac{d \phi}{d t}\right|$ $e=\frac{\mu_{0} \cdot \pi a^{2}}{2 b} \dot{i}_{0} \cdot \omega \sin \omega t$
Hence, the correct option is B.
Question 8
1 / -0

Four wire A, B, C and D each of length $\ell=10 \mathrm{cm}$and each of area of cross section is 0.1 m2 are connected in the given circuit. Then, the position of null point is

Given that resistivity

$\rho_{A}=1 \Omega-m$
$\rho_{B}=3 \Omega-m$
$\rho_{C}=6 \Omega-m$
$\rho_{D}=1 \Omega-m$

A
mid point of wire B or wire C when both the switches, S₁ and S₂ open.

B
only mid point of wire B when both the switches, S₁ and S₂ are open.

C
mid point of wire D when both the switches S₁ and S₂ are open.

D
3 cm from the left end of B when both the switches S₁ and S₂ are open.

Solution

When S1 and S2 both open and jockey has no deflection then let current in upper circuit is ” i”.$\mathrm{R}=\frac{\delta \mid}{\mathrm{A}}$

So, $\mathrm{R}_{\mathrm{A}}=10$

$\mathrm{R}_{\mathrm{B}}=30$

$\mathrm{R}_{\mathrm{C}}=6 \Omega$

$\mathrm{R}_{\mathrm{D}}=10$

$\mathrm{R}_{\mathrm{eq}}=4+1+1+\frac{3.6}{3+6}=8 \Omega$

$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{8}{8}=1$ Ampere

When jockey is touched, and there is no deflection then potential difference between ‘A’ and the point should be same in both path-

So, in lower circuit as S2 is open

No current will flow in this circuit

⇒Potential difference between A and P

Will be = 2V

⇒i1 . 3= i2 . 6 and i1 + i2 = 1

$\Rightarrow \mathrm{i}_{1}=\frac{2}{3} \mathrm{A}, \quad \mathrm{i}_{2}=\frac{1}{3} \mathrm{A}$
Potential difference between A and P
$\Rightarrow \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{p}}=\mathrm{i} .1+\mathrm{i}_{2} \cdot\left(\mathrm{R}^{\prime}\right)$
$\Rightarrow 2=1+R^{\prime}\left(\frac{1}{3}\right)$
$\Rightarrow \quad \mathbb{R}^{\prime}=3$
For this point ‘P’ should be half way in wire C.
Similarly when jockey touched on wire ‘B’, Null point will be obtained at middle point of wire B.
Hence, the correct option is A.
Question 9
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Three identical bulbs each of resistance 2Ω are connected as shown. The maximum power that can be consumed by individual bulb is 32W, then the maximum power consumed by the combination is :

A
48 W

B
96 W

C
128 W

D
160 W

Solution

Resistance of B1, B2 and B3 are same which is = 2Ω

As P ∝i2 (when ‘R’ constant)
So, maximum power consumed by bulb B3
Which is = 32 W
⇒i2R = 32
⇒i2 . (2) = 32
⇒i = 4 Ampere
So, current passing through B1 + B2 = i/2 = 2
Total power consumed in circuit
P = (i/2)2 (2) + (i/2)2 (2) + i2(2)
= (2)2 (2) + (2)2 (2) + (4)2 (2)
= 48W
Hence, the correct option is A.
Question 10
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A car completes a certain journey in 8 hours. It covers half the distance at 40 km/hr and the rest at 60 km/hr. The length of the journey is.....

A
350 km

B
420 km

C
384 km

D
400 km

Solution

Let the total journey be $x$ km.
$\frac{x}{2} \times \frac{1}{40}+\frac{x}{2} \times \frac{1}{60}=8$
Therefore, $\frac{x}{8}+\frac{x}{120}=8$
$3 x+2 x=1920$
That is, $5 x=1920, x=384 km$
Hence, the correct option is C.