Class 12th

An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0g cm^-3. The number of atoms in 180g of the element is_________* 10²³ (Nearest integer)

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Contributor-Level 10

B.C.C structure

a = 300 pm = 300 * 10^-12 m

d = 6g/cm³

z = 2

$d = \frac{Z * M}{a^{3}}$

$6 = \frac{2 * A}{{(300 * 1 0^{- 10})}^{3}} = \frac{2 * A}{27 * 1 0^{- 24}}$

$∴ A t o m s o f M$ = 3.69 * 6.022 * 10²³

= 22.22 * 10²³

the nearest integer = 22

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6 months ago

Two electric dipoles of dipole moments 1.2 * 10^-³⁰ Cm and 2.4 * 10^-³ Cm are placed in two different uniform electric fields of strengths 5 * 10⁴ NC^-¹ and 15 * 10⁴ NC^-¹ respectively. The ratio of maximum torque experienced by the electric dipoles will be $\frac{1}{x} .$ The value of x is_______.

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$P_{1} = 1.2 * 1 0^{- 30} C . m P_{2} = 2.4 * 1 0^{- 30} C . m$

$E_{1} = 5 * 1 0^{4} N C^{- 1} E_{2} = 15 * 1 0^{4} N C^{- 1}$

So, $\frac{τ_{1}}{τ_{2}} = \frac{P_{1} E_{1} s i n 90 °}{P_{2} E_{2} s i n 90 °} = \frac{(1.2 * 1 0^{- 30}) * (5 * 1 0^{4})}{(2.4 * 1 0^{- 30}) * (15 * 1 0^{4})}$

$= \frac{1}{2} * \frac{1}{3} = \frac{1}{6} \Rightarrow x = 6$

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As shown in the figure, a potentiometer wire of resistance $20 Ω$ and length 300cm is connected with resistance box (R.B) and a standard cell of emf 4 V. For a resistance 'R' of resistance box introduced into the circuit, the null point for a cell of 20 mV is found to be 60 cm. The value of 'R' is __________ $Ω .$

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Contributor-Level 10

$R_{A B} = 20 Ω$

L = 300 cm

Null point is at 60 cm from A, so

$20 m V = \frac{4}{R + 20} * 20 * (\frac{60}{300}) * 1 0^{3}$

$\Rightarrow 20 = \frac{16}{R + 20} * 1 0^{3}$

$R = \frac{15600}{20} = 780 Ω$

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The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{\sqrt[]{2}}$ times its maximum value, are 212 rad s^-¹ and 232 rad^-¹. The value of resistance in the circuit is R = $5 Ω .$ The self inductance in the circuit is_________mH.

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Contributor-Level 10

$\frac{_{}}{}$ $i = \frac{l_{0}}{\sqrt[]{2}}$ at $ω_{1}$ $_{}$ = 212 rad/s $R = 5 Ω$

$_{}$ $ω_{2}$ = 232 rad/s $Δ ω = ω_{2} - ω_{1} = 20 r a d / s$ $_{}_{}$

$P = \frac{P_{m a x}}{2} a t ω_{1} a n d ω_{2}$

So, $Δ ω = ω_{2} - ω_{1} = \frac{R}{L}$ $_{}_{} \frac{}{}$

$\Rightarrow L = \frac{R}{Δ ω} = \frac{5}{20} = . 25 H$

= 250 mH

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Class 12th Wave Optics

In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is__________nm.

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Contributor-Level 10

$λ_{1} = 560 n m,$ fringe width, B = 7.2 mm

$\Rightarrow B_{1} = \frac{D λ_{1}}{d} = 7.2 m m$

For $λ_{2} \to B_{2} = \frac{D λ_{2}}{d} = 8.1 m m$ $_{}_{} \frac{_{}}{}$

$\frac{B_{2}}{B_{1}} = \frac{λ_{2}}{λ_{1}} = \frac{8.1}{7.2} \Rightarrow λ_{2} \frac{81}{72} λ_{1}$

$= \frac{9}{8} * 560 n m$

= 630 nm

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A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be __________hours.

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Contributor-Level 10

t_1/2= 2hr 30 min

Radiation intensity a disintegrations / sec (Activity)

As, $A = \frac{A_{0}}{{2^{t}}^{\frac{t}{2}}}$

For safe working, $A = \frac{A_{0}}{64} \Rightarrow \frac{A_{0}}{64} = \frac{A_{0}}{2^{\frac{t}{t_{\frac{1}{2}}}}}$

$\Rightarrow t = 6 t_{\frac{1}{2}} = 6 * 2.5 h r$

= 15

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Number of grams of bromine that will completely react with 5.0g of pent-1-ene is_____________*10^-²g. (Atomic mass of Br = 80 g/mol) (Nearest Integer)

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Payal Gupta

Contributor-Level 10

Moles of Br₂= moles of C₅H₁₀= $\frac{5}{60 + 10} = \frac{5}{70}$

$∴ \frac{w}{160} = \frac{5}{70}$

$∴ w = \frac{5 * 160}{70} = \frac{80}{7} g$

$= 1142.8 * 1 0^{- 2} \approx 1143 * 1 0^{- 2} g$

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If ${[C u {(H_{2} O)}_{4}]}^{2 +}$ absorbs a light of wavelength 600nm d-d transition, then the value of octahedral crystal field splitting energy for ${[C u {(H_{2} O)}_{6}]}^{2 +}$ will be_________* 10^-^21J. (Nearest Integer)

(Given: h = 6.63 * 10^-³⁴ Js and c = 3.08 * 10⁸ ms^-¹)

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Payal Gupta

Contributor-Level 10

$Δ_{t} = \frac{h c}{λ} = \frac{6.63 * 1 0^{- 34} * 3.08 * 1 0^{8}}{600 * 1 0^{- 9}}$

$= \frac{6.63 * 3.08 * 1 0^{- 17}}{600}$ J

$= 765.765 * 1 0^{- 21} J$

$? 766 * 1 0^{- 21}$ J

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