Physics

A thin lens made of glass (refractive index $= 1.5$ ) of focal length $f = 16 c m$ is immersed in a liquid of refractive index 1.42. If its focal length in liquid is $f_{l}$ , then the ratio $f_{l}$ /f is closest to the integer:

0 Follower 3 Views

Contributor-Level 10

$\frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f_{l}} = (\frac{μ_{g}}{μ_{l}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

(i) $\frac{f_{l}}{f} = \frac{μ_{g} - 1}{\frac{μ_{g}}{μ_{l}} - 1} = \frac{1.5 - 1}{\frac{1.5}{1.41} - 1} = 8.875 \approx 9$

0 0

New answer posted

11 months ago

Time taken by light to travel in two different materials A and B refractive indices $μ_{A} a n d μ_{B}$ of same thickness is $t_{1} a n d t_{2}$ respectively. It $t_{2} - t_{1} = 5 * 1 0^{- 10} s$ and the ratio of $μ_{A} t o μ_{B}$ 1 : 2. Then, the thickness of material, in meter is: (Given $υ_{A} a n d υ_{B}$ are velocities of light in A and B materials respectively.)

0 Follower 3 Views

Contributor-Level 10

$\frac{μ_{A}}{μ_{B}} = \frac{C / v_{A}}{C / v_{B}} = \frac{v_{B}}{v_{A}} = \frac{1}{2}$

Let the thickness is d

$d = \frac{5 * 1 0^{- 10} * v_{A} v_{B}}{v_{A} - v_{B}}$

d = 5 * 10^-10 * v_A

0 0

New answer posted

11 months ago

Which of the following statement is correct?

0 Follower 1 View

Physics Dual Nature of Radiation and Matter

Contributor-Level 10

In primary rainbow, red colour is at top and violet is at bottom.

Intensity of secondary rainbow is less in comparison to primary rainbow.

0 0

New answer posted

11 months ago

An electron (of mass $m$ ) and a photon have the same energy $E$ in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ( $c =$ speed of light in vacuum)

0 Follower 7 Views

Contributor-Level 10

$λ_{electron} = \frac{h}{\sqrt[]{2 m E}} \frac{λ_{electron}}{λ_{photon}} = \frac{1}{c} \sqrt[]{\frac{E}{2 m}}$

$λ_{photon} = \frac{h c}{E}$

0 0

New answer posted

11 months ago

The rms value of conduction current in a parallel plate capacitor is 6.9 $μ A$ . The capacity of this capacitor, if it is connected to 203 V ac supply with an angular frequency of 600 rad/s, will be:

0 Follower 5 Views

Contributor-Level 10

Current in capacitor l = $\frac{V}{X_{C}}$ $\frac{}{_{}}$

I = v * $(ω c)$

$C = \frac{1}{v ω} = \frac{6.9 * 1 0^{- 6}}{230 * 600} = 50 p F$

0 0

New answer posted

11 months ago

A small square loop of wire of side $l$ is placed inside a large square loop of wire. L(L >> $l$ ). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is:

0 Follower 17 Views

Contributor-Level 10

Assuming current 'i' In outer loop magnetic field at centre

$= 4 * \frac{μ_{0}}{4 π} \frac{i}{L / 2} * [2 s i n 45 °)$

$= \frac{2 \sqrt[]{2} μ_{0} i}{π L}$

M = $\frac{F m α t h r o u g h i n n e r l o o p}{i}$ $\frac{}{}$

= $\frac{2 \sqrt[]{2} μ_{0} i l^{2}}{π L} = \frac{2 \sqrt[]{2} μ_{0} l^{2}}{π L}$ $\frac{_{}^{}}{} \frac{_{}^{}}{}$

0 0

New answer posted

11 months ago

In a building there are 15 bulbs of $45 W, 15$ bulbs of $100 W, 15$ small fans of $10 W$ and 2 heaters of $1 k W$ . The voltage of electric main is $220 V$ . The minimum fuse capacity (rated value) of the building will be

0 Follower 2 Views

Contributor-Level 10

Total power is $(15 * 45) + (15 * 100) + (15 * 10) + (2 * 1000) = 4325 W$

So, current is $\frac{4325}{220} = 19.66 A$

Answer is $20 A m p$ .

0 0

New answer posted

11 months ago

To increase the resonant frequency in series LCR circuit,

0 Follower 2 Views

Physics Physics Moving Charges and Magnetism

Contributor-Level 10

$f = \frac{1}{2 π \sqrt[]{L C}}$

$f α \frac{1}{\sqrt[]{C}}$

Another capacitor should be added in series with the first capacitor.

0 0

New answer posted

11 months ago

A particle of mass $m$ and charge $q$ has an initial velocity $\vec{v} = v_{0} \hat{j}$ . If an electric field $\vec{E} = E_{0} \hat{i}$ and magnetic field $\vec{B} = B_{0} \hat{i}$ act on the particle, its speed will double after a time:

0 Follower 16 Views

Physics Physics Moving Charges and Magnetism

Contributor-Level 10

Since magnetic force cannot change the speed. So only electric field which is along -direction will change the speed along -direction only.

$v_{x} = \frac{E_{0} q}{m} t$ but $v_{y} = v_{0}$

$2 v_{0} = \sqrt[]{v_{x}^{2} + v_{y}^{2}}$

4 v_{0}^{2} = \frac{E_{o}^{2} q^{2} t^{2}}{m^{2}} + v_{0}^{2}

t = \frac{\sqrt[]{3} m v_{o}}{q E_{o}}

0 0

New answer posted

11 months ago

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. It the ratio of radii of their circular paths is 6 : 5 and their respective masses ratio is 9 : 4. Then, the ratio of their charges will be:

0 Follower 2 Views