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Physics Ncert Solutions Class 12th

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Physics Ncert Solutions Class 12th Class 12th

A loop ABCDEFA of straight edges has six corner points $A (0,0, 0), B (5,0, 0), C (5,5$ , $0), D (0,5, 0), E (0,5, 5)$ and $F (0,0, 5)$ . The magnetic field in this region in $\overset{?}{B} = (3 \hat{i} + 4 \hat{k}) T$ . The quantity of flux through the loop ABCDEFA (in Wb) is

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Vishal Baghel

Contributor-Level 10

$? = \overset{?}{B} \cdot \overset{?}{A} = (3 \hat{i} + 4 \hat{k}) \cdot (25 \hat{i} + 25 \hat{k})$

$? = (3 * 25) + (4 * 25) = 175 weber$

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Physics Ncert Solutions Class 12th Class 12th

The number of chlorine atoms in bithionol is_____________.

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alok kumar singh

Contributor-Level 10

Structure of Bithinol is;

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Physics Ncert Solutions Class 12th Class 12th

A capacitor of capacitance 500 $μ F$ is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.

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alok kumar singh

Contributor-Level 10

C = 500 $μ F,$ V = 100 v, L = 50 mH

In this LC – oscillation

q = q₀ cos $ω t$

$i = \frac{- d q}{d t} = q_{0} ω s i n ω t ω = \frac{1}{\sqrt[]{2 c}} = \frac{1}{\sqrt[]{50 * 1 0^{- 3} * 5 * 1 0^{- 4}}}$

= $\frac{1000}{5} = 200$

So, i_max = $q_{0} ω = 500 * 1 0^{6} * 100 * 200$

= 10A

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Physics Ncert Solutions Class 12th Class 12th

A 8 V Zener diode along with a series resistance R is connected across a 20V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be_______ $Ω$ .

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alok kumar singh

Contributor-Level 10

$i_{(z e n e r - m a x)} = 25 m A$

20 – i_maxR – 8 = 0

i_max R = 12At minimum zener current $(μ A) : -$

20 - i_{m i n} R - i_{m i n} R_{L} = 0

$\frac{R}{R_{L}} = \frac{12}{8} = \frac{3}{2}$

$l_{m i n} R = 12$

$i_{m i n} R_{L} = 8$

At max^m zener current –

$20 - i_{m a x} R - 8 = 0$

$i_{L} = O {a s i_{z} m a x^{m} = 25 m A}$

i_maxR = 12v

25 * 10^-3 R = 12

$R = \frac{12 * 1 0^{3}}{25} = 12 * 40 = 480 Ω$

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Physics Ncert Solutions Class 12th Class 12th

A modulating signal 2sin (6.28 * 10⁶)t is added to the carrier signal 4 sin(12.56 *10⁹)t for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passes through a band pass filter. The bandwidth of the output signal of band pass filter will be_________ MHz.

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alok kumar singh

Contributor-Level 10

Modulating signal 2sin (6.28 * 10⁶)t

Carrier signal 4 sin (12.56 * 10⁹)t

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Physics Ncert Solutions Class 12th Class 12th

A conducting circular loop is placed in X – Y plane in presence of magnetic field $\vec{B} = (3 t^{3} \hat{j} + 3 t^{2} \hat{k})$ in SI unit. If the radius of the loop is 1m. The induced emf in the loop, at time, t = 2s is npV. The value of n is__________.

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Vishal Baghel

Contributor-Level 10

$? = \vec{B} . \vec{A}$

$= (3 t^{3} \hat{j} + 3 t^{2} \hat{k}) . (π {(1)}^{2} \hat{k})$

$? = 3 t^{2} π$

$E_{i n d} = | \frac{d ?}{d t} | = 6 t π$

at t = 2, E_ind = 12

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Physics Ncert Solutions Class 12th Class 12th

A thin prism of angle 6° and refractive index for yellow light (n_Y) 1.5 is combined with another prism of angle 5° and n_Y = 1.55. The combination produces no dispersion. The net average deviation $(δ)$ produced by the combination is ${(\frac{1}{x})}^{o}$ . The value of x is____________.

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Vishal Baghel

Contributor-Level 10

$δ = A (μ_{y} - 1) - A^{1} (μ_{y}^{1} - 1)$

$= 6 (1.5 - 1) - 5 (1.55 - 1)$

$= \frac{1}{4}$

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Physics Ncert Solutions Class 12th 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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Payal Gupta

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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Physics Ncert Solutions Class 12th Class 12th

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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Payal Gupta

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