Class 12th

Magnetic flux (in weber) in a closed circuit of resistance $20 Ω$ varies with time t(s) as $?$ = 8t² – 9t + 5. The magnitude of the induced current at t = 0.25s will be _______mA.

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Contributor-Level 10

$e_{i n} = | \frac{d ?}{d t} | = | 16 t - 9 | e_{i n} = | 16 * \frac{1}{4} - 9 | = 5 v o l t$

$l = \frac{e_{i n}}{R} = \frac{5}{20} A = \frac{5}{20} * 1000 m A = 250 m A$

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11 months ago

A convex lens of focal length 20cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is ______cm.

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Class 12th Physics Electrostatic Potential and Capacitance

Contributor-Level 10

For refraction of light at lens

$\frac{1}{v} - \frac{1}{- 60} = \frac{1}{20}$

$\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{60} \Rightarrow \frac{1}{v} = \frac{3 - 1}{60} = \frac{1}{30}$

v = 30 cm

If image has to be formed at object itself then light ray should retrace its path. Hence after refraction at lens, it must strikes normally to the mirror

RM = 20 cm Radius of curvature of mirror

Fm = 10 cm focal length of mirror

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11 months ago

Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 F. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ________V.

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Contributor-Level 10

$\frac{C V + q}{9 C} + \frac{- 3 C V + q}{3 C} = 0$

$\Rightarrow C V + q - 9 C V + 3 q = 0$

$\Rightarrow q = \frac{8 C V}{4} = 2 C V$

$V_{A B} = \frac{C V + q}{9 C} = \frac{3 C V}{9 C} = \frac{V}{3} = \frac{18}{3} = 6 v o l t$

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11 months ago

Two ideal diodes are connected in the network as shown is figure. The equivalent resistance between A and B is _______ $Ω$ .

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Contributor-Level 10

Here diode-D₁ is forced biased and diode-D₂ is reversed biased, so

$R_{A B} = (20 / / 20) + 15 = 25 Ω$

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11 months ago

In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ________cm.

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Contributor-Level 10

$V = K l \Rightarrow \frac{V_{2}}{V_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{l_{2}}{36} = \frac{1.80}{1.20} = \frac{3}{2} \Rightarrow l_{2} = 54 c m$

$Δ l = l_{2} - l_{1} = 54 - 36 = 18 c m$

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11 months ago

Class 12th Physics Atoms

$\frac{x}{x + 4}$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its

(i) third permitted energy level to the second level and

(ii) the highest permitted energy level to the second permitted level.

The value of x will be ________.

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Contributor-Level 10

According to definition of atomic energy, we can write

$E_{1} = E_{0} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 E_{0}}{36}, a n d E_{2} = E_{0} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$

$\frac{E_{1}}{E_{2}} = \frac{5}{9} \Rightarrow \frac{x}{x + 4} = \frac{5}{5 + 4} \Rightarrow x = 5$

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11 months ago

Class 12th Maths

The line of shortest distance between the lines $\frac{x - 2}{0} = \frac{y - 1}{1} = \frac{z}{1}$ and $\frac{x - 3}{2} = \frac{y - 5}{2} = \frac{z - 5}{1}$ makes an angle of $c o s^{- 1} (\sqrt[]{\frac{2}{27}})$ with the plane P : ax – y – z = 0, (a > 0). If the image of the point (1, 1, - 5) in the plane P is $(α, β, γ),$ then $α + β - γ$ is equal to……….

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Contributor-Level 10

Line of shortest distance will be along ${\bar{b}}_{1} * {\bar{b}}_{2}$ $_{}_{}$

where $\begin{array}{l} _{} \end{array}$ $\begin{array}{l} {\bar{b}}_{1} = \hat{j} + \hat{k} \\ a n d \\ {\bar{b}}_{2} = 2 \hat{i} + 2 \hat{j} + \hat{k} \\ {\bar{b}}_{1} * {\bar{b}}_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 2 & 2 & 1 \end{matrix} | = - \hat{i} + 2 \hat{j} - 2 \hat{k} \end{array}$

Angle between ${\bar{b}}_{1} * {\bar{b}}_{2}$ $_{}_{}$ and plane P,

$\Rightarrow a^{2} = - \frac{25}{11} (n o t p o s s i b l e)$

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11 months ago

Class 12th Maths

If $\underset{n \to \infty}{l i m} \frac{{(n + 1)}^{k - 1}}{n^{k + 1}} [(n k + 1) + (n k + 2) + . . . . + (n k + n)]$

$= 33 . \underset{n \to \infty}{l i m} \frac{1}{n^{k + 1}} .$ $[1^{k} + 2^{k} + 3^{k} + . . . + n^{k}]$ , then the integral value of k is equal to…………

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Contributor-Level 10

$l i m_{n \to \infty} {(\frac{n + 1}{n})}^{k - 1} \frac{1}{n} \sum_{r = 1}^{n} (k + \frac{r}{n})$

= $33 \int_{0}^{1} x^{k} d x$

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11 months ago

Class 12th Maths

Let f(x) = ${\begin{matrix} | 4 x^{2} - 8 x + 5 |, i f 8 x^{2} - 6 x + 1 \geq 0 & [4 x^{2} - 8 x + 5], i f 8 x^{2} - 6 x + 1 < 0 \end{matrix}}$ , where [a] denotes the greatest integer less than or equal to a. Then the number of points in R where f is not differentiable is………….

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Contributor-Level 10

$f (x) = {\begin{cases} | 4 x^{2} - 8 x + 5 |, i f 8 x^{2} - 6 x + 1 \geq 0 \\ [4 x^{2} - 8 x + 5], i f 8 x^{2} - 6 x + 1 < 0 \end{cases}$

$x = \frac{1}{4}, \frac{2 - \sqrt[]{2}}{2}, \frac{1}{2}$

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11 months ago

Class 12th Binomial Theorem

If the maximum value of the term independent of t in the expansion of ${(t^{2} x^{\frac{1}{5}} + \frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{15}$ , x $\geq$ , is K, then 8 K is equal to………….

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