WBJEE 2016 Paper 2 (Physics & Chemistry) Question Paper with Solutions

$\frac{1}{𝐶}=\frac{1}{{𝐶}_1}+\frac{1}{{𝐶}_2}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$

C = 2$𝜇𝐹$

2$𝜇𝐹$ will be parallel to 4$𝜇𝐹$, so the resultant capacitance



${𝐶}^{\prime }={𝐶}_1+{𝐶}_2=2𝜇𝐹+4𝜇𝐹$

$=6𝜇𝐹$



$\frac{1}{{𝐶}^{{}^{\prime \prime }}}=\frac{1}{4𝜇𝐹}+\frac{1}{4𝜇𝐹}=\frac{2}{4}𝜇𝐹$

${𝐶}^{\prime \prime }=2𝜇𝐹$

$\therefore$ ${𝐶}_{𝐴𝐵}=2𝜇𝐹+6𝜇𝐹$

$=8𝜇𝐹$

$𝑅=\frac{𝜌𝑙}{𝐴}$

where, $𝜌$ = resistivity

l = length of the resistance wire

A = area of cross-section

When the wires are connected in series, then

$𝑅={𝑅}_1+{𝑅}_2$

$𝑅=\frac{{𝜌}_1{𝑙}_1}{{𝐴}_1}+\frac{{𝜌}_1{𝑙}_2}{{𝐴}_2}$

$=\frac{{𝜌}_1{𝑙}_1}{𝐴}+\frac{{𝜌}_2{𝑙}_2}{𝐴}$ $\because$ $({𝐴}_1={𝐴}_2)$

${𝜌}_{𝑒𝑞}$ = resistivity of combination

$𝑙={𝑙}_1+{𝑙}_2$

$𝐴=𝐴$

Thus, ${𝜌}_{𝑒𝑞}.\frac{{𝑙}_1+{𝑙}_2}{𝐴}=\frac{{𝜌}_1{𝑙}_1+{𝜌}_2{𝑙}_2}{𝐴}$

${𝜌}_{𝑒𝑞}=\frac{{𝜌}_1{𝑙}_1+{𝜌}_2{𝑙}_2}{{𝑙}_1+{𝑙}_2}$

Inside a hollow charged sphere, the electric field intensity is always zero.

$\therefore$ E = 0

but potential is constant.



Potential on the surface of the hollow sphere

$𝑉=\frac{𝐾𝑄}{𝑅}=\frac{1}{4𝜋{𝜀}_0}.\frac{𝑄}{𝑅}=\frac{𝑄}{4𝜋{𝜀}_0𝑅}$

This is the potential (V) inside the charged hollow metal sphere.

The de-Broglie wavelength associated with electron is given by,

$𝜆=\frac{ℎ}{\sqrt{\surd 2𝑚𝑒𝑉}}$

where,

h = Planck's constant

m = mass of electron

e = charge on electron

V = potential difference

$𝜆=1\stackrel{𝑜}{𝐴}={10}^{-10}𝑚$ (given)

${𝜆}^2=\frac{{ℎ}^2}{2𝑚𝑒𝑉}=𝑉=\frac{{ℎ}^2}{2𝑚𝑒{𝜆}^2}$

$=\frac{{(6.6\times {10}^{-34})}^2}{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times {(1\times {10}^{-10})}^2}$

$=\frac{6.6\times {10}^{-34}\times 6.6\times {10}^{-34}}{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times {10}^{-20}}$

$=\frac{6.6\times 6.6\times {10}^2}{2\times 9.1\times 1.6}=\frac{4356\times {10}^2}{2912}$ = 150 V

Work function of cesium = 2.27 eV

The wavelength of incident light

$𝜆$ = 600 nm = 600 $\times$ 10^$-$9 m

The energy of incident photon = hv

$𝐸=\frac{ℎ𝑐}{𝜆}$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{600\times {10}^{-9}\times 1.6\times {10}^{-19}}$ eV

= 2.06 eV

$\frac{ℎ𝑐}{𝜆}$ < $𝜙$ (work function)

2.06 eV < 2.27 eV

So, no emission of electrons will take place.

In the Bohr model of the hydrogen atom, the quantization condition for the angular momentum states that the angular momentum is an integer multiple of $\hslash$ (reduced Planck's constant). Mathematically, this is given by:

$𝑚𝑣𝑟=𝑛\hslash$

where:

$𝑚$ is the electron mass,

$𝑣$ is the electron velocity,

$𝑟$ is the radius of the orbit,

$𝑛$ is the principal quantum number,

$\hslash =\frac{ℎ}{2𝜋}$ (where $ℎ$ is Planck's constant).

According to de-Broglie, an electron in a stable orbit behaves like a standing wave. The condition for standing waves is that the circumference of the orbit must be an integer multiple of the electron's de-Broglie wavelength $𝜆$, which is given by:

$𝜆=\frac{ℎ}{𝑚𝑣}$

For the wave to be stable, the circumference of the orbit $2𝜋𝑟$ must equal $𝑛$ times the de-Broglie wavelength:

$2𝜋𝑟=𝑛𝜆$

Substituting for $𝜆$ gives:

$2𝜋𝑟=𝑛\left(\frac{ℎ}{𝑚𝑣}\right)$

This equation is consistent with the quantization condition. It shows that the number of de-Broglie wavelengths that can fit in the orbit is equal to the principal quantum number $𝑛$.

For the second Bohr orbit, $𝑛=2$. Hence, the number of de-Broglie wavelengths contained in the second Bohr orbit is:

Option B: 2

$\frac{1}{𝜆}=𝑅\left[\frac{1}{{𝑛}_2^1}+\frac{1}{{𝑛}_2^2}\right]$

For Balmer series,

n₁ = 2, n₂ = 3, 4, 5 .....

$\therefore$ $\frac{1}{{𝜆}_{𝐵}}=𝑅\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$

$𝑅=\left[\frac{1}{4}-\frac{1}{16}\right]$

$\frac{1}{{𝜆}_{𝐵}}=𝑅.\frac{3}{16}$ ..... (i)

Lyman series is obtained when n₁ = 1, n₂ = 2, 3, 4, ......

For third line in Lyman series,

$\frac{1}{{𝜆}_{𝐿}}=𝑅\left[\frac{1}{1^2}-\frac{1}{4^2}\right]=𝑅\left[1-\frac{1}{16}\right]$

$\frac{1}{{𝜆}_{𝐿}}=𝑅.\frac{15}{16}$ ...... (ii)

$\frac{1/{𝜆}_{𝐵}}{\frac{1}{{𝜆}_{𝐿}}}=\frac{𝑅.\frac{3}{16}}{𝑅.\frac{15}{16}}$

$\frac{{𝜆}_{𝐿}}{{𝜆}_{𝐵}}=\frac{3}{15}$ ..... (iii)

${𝜆}_{𝐿}=\frac{3}{15}\times {𝜆}_{𝐵}$

${𝜆}_{𝐵}$ = 600 nm = 600 $\times$ 10^$-$9 m

${𝜆}_{𝐿}=\frac{3}{15}\times 600\times {10}^{-9}$

= 120 $\times$ 10^$-$9 m = 120 nm.

Here, angle of incidence is 60${}^{\circ }$



$𝜇=\frac{\sin 𝑖}{\sin 𝑟}$

$\sin 𝑖=𝜇\sin 𝑟$

here, $𝑟=90-𝑖$

$\sin 𝑖=𝜇\sin ({90}^{\circ }-𝑖)=𝜇\cos 𝑖$

$𝜇=\frac{\sin 𝑖}{\cos 𝑖}=\tan 𝑖=\tan {60}^{\circ }$

$𝜇=\sqrt{\surd 3}$

Thickness of glass plate = t

Refractive index = $𝜇$

Velocity of light = c

Speed of light = $\frac{𝑑𝑖𝑠\tan 𝑐𝑒}{𝑡𝑖𝑚𝑒}=\frac{𝑡}{𝑇}$

T = $\frac{𝑡}{𝑠𝑝𝑒𝑒𝑑𝑜𝑓𝑙𝑖𝑔ℎ𝑡}$

Speed of light in glass plate

$𝑣=\frac{𝑐}{𝜇}$ ($\because$ $𝜇=\frac{𝑐}{𝑣}$)

$𝑇=\frac{𝑡}{𝑐/𝜇}=\frac{𝜇𝑡}{𝑐}$

Given, x = at + bt²

[ x ] = [ bt² ]

Unit of b = $\frac{𝑥}{{𝑡}^2}=\frac{𝑚𝑒𝑡𝑟𝑒}{{\left(ℎ\right)}^2}=\frac{𝑚}{{ℎ}^2}$

Given, | A + B | = | A $-$ B |

$\therefore$ (A + B)² = (A $-$ B)²

A² + B² + 2AB cos $𝜃$ = A² + B² $-$ 2 AB cos $𝜃$

2 AB cos $𝜃$ + 2 AB cos $𝜃$ = 0

4 AB cos $𝜃$ = 0

Here, A $\ne$ 0, B $\ne$ 0

$\therefore$ cos $𝜃$ = 0

or $𝜃$ = 90${}^{\circ }$

Given that velocity of ascending body = u



Let there is no air resistance when body falls freely in downward direction, the direction of falling body will opposite to the ascending body.

$\therefore$ The velocity at the same height while the body falls freely = $-$ u

The bodies are separated by a distance R, so the coordinates of m₁ and m₂ will be (0, 0) and (R, 0)



From the formula of centre of mass,

${𝑋}_{𝑐𝑚}=\frac{{𝑚}_1{𝑥}_1+{𝑚}_2{𝑥}_2}{{𝑚}_1+{𝑚}_2}$

$=\frac{{𝑚}_1\times 0+{𝑚}_2\times 𝑅}{{𝑚}_1+{𝑚}_2}$

${𝑋}_{𝑐𝑚}=\frac{{𝑚}_2.𝑅}{{𝑚}_1+{𝑚}_2}$

T₁ = 273 + 20 = 293 K

T₂ = 273 + 40 = 313 K

The velocity of sound wave in gases or air is given by

$𝑣=\sqrt{\surd \frac{𝛾𝑅{𝑇}_1}{𝑀}}$

where, $𝛾$ = Ratio of C_p/C_v

R = gas constant

$\therefore$ ${𝑣}_1=\sqrt{\surd \frac{𝛾𝑅{𝑇}_1}{𝑀}}$

${𝑣}_2=\sqrt{\surd \frac{𝛾𝑅{𝑇}_2}{𝑀}}$

$\frac{{𝑣}_1}{{𝑣}_2}=\sqrt{\surd \frac{{𝑇}_1}{{𝑇}_2}}=\sqrt{\surd \frac{293}{313}}$

Given, v₁ = 344.2 m/s

$\therefore$ ${𝑣}_2={𝑣}_1\sqrt{\surd \frac{313}{293}}$

$=344.2\times \sqrt{\surd 1068}$

$=344.2\times 103$ m/s

$=355.75$ m/s

$\simeq 356$ m/s

Mass of gas = 4g (given)

From perfect gas equation

pV = nRT

where n = number of moles

here, $\left(𝑛=\frac{4}{2}=2\right)$

$\therefore$ pV = 2RT

It will be the perfect gas equation for 4g of hydrogen.

Suppose temperature of Sun = T

When it is doubled, T' = 2T

From law of radiation (Stefan's law)

E $\propto$ T⁴ = $𝜎$T⁴

$𝜎$ = Stefan's constant.

E₁ $\propto$ T⁴

E₂ $\propto$ (2T)⁴

$\frac{{𝐸}_1}{{𝐸}_2}=\frac{{𝑇}^4}{2^4.{𝑇}^4}$

E₂ = 16 E₁

In SHM, the acceleration of vibrating particle is proportional to displacement

| a | = $𝜔$²x.

Here, a = 16 cm/s²

= 16 $\times$ 10^$-$2 m/s²

Displacement, x = 4 cm = 4 $\times$ 10^$-$2 m

$\therefore$ 16 $\times$ 10^$-$2 = $𝜔$² (4 $\times$ 10^$-$2)

$𝜔$² = 4

$𝜔$ = $\pm$ 2 rad/s

$𝜔$ = $\frac{2𝜋}{𝑇}$

T = $\frac{2𝜋}{𝜔}$ = $\frac{2𝜋}{2}$ = $𝜋$ = 3.142 s

The work done in stretching the spring, $𝑊=\frac{1}{2}𝐾{𝑥}^2$

K = force constant of spring

x = extension in length of the spring

${𝑊}_1=\frac{1}{2}𝐾{𝑥}_2^1$

${𝑊}_2=\frac{1}{2}𝐾{𝑥}_2^2$

${𝑥}_1=1$

${𝑥}_2=1+1=2$

$\frac{{𝑊}_1}{{𝑊}_2}=\frac{\frac{1}{2}𝐾{\left(1\right)}^2}{\frac{1}{2}𝐾{\left(2\right)}^2}=\frac{1}{4}$

${𝑊}_2=4{𝑊}_1$

${𝑊}_1=10𝐽$

${𝑊}_2=40𝐽$

$\therefore$ More work required = $40𝐽-10𝐽=30𝐽$

${𝑣}_{𝑟𝑚𝑠}=\sqrt{\surd \frac{3𝑅𝑇}{𝑀}}$

R = gas constant

T = absolute temperature

M = mass of gas

${𝑣}_{𝑟𝑚𝑠}$ of ${𝐻}_2=\sqrt{\surd \frac{3𝑅𝑇}{{𝑀}_{{𝐻}_2}}}$

${𝑣}_{𝑟𝑚𝑠}$ of ${𝑂}_2=\sqrt{\surd \frac{3𝑅𝑇}{{𝑀}_{{𝑂}_2}}}$

$\frac{{𝑣}_{𝑟𝑚𝑠}.{𝐻}_2}{{𝑣}_{𝑟𝑚𝑠}.{𝑂}_2}=\sqrt{\surd \frac{{𝑀}_{{𝑂}_2}}{{𝑀}_{{𝐻}_1}}}$

Given, ${𝑣}_{𝑟𝑚𝑠}{𝐻}_2=𝐶$

$\Rightarrow \frac{𝐶}{{𝑣}_{𝑟𝑚𝑠}{𝑂}_2}=\sqrt{\surd \frac{{𝑀}_{{𝑂}_2}}{{𝑀}_{{𝐻}_1}}}=\sqrt{\surd \frac{32}{2}}=\sqrt{\surd 16}=4$

$\therefore$ ${𝑣}_{𝑟𝑚𝑠}=\frac{𝑐}{4}$

The tube is closed at one end. So, it will act like a closed organ pipe.

The lowest frequency produced in the tube



l = $\frac{𝜆}{4}$, $𝜆$ = 4

l = 1 m

f = 84 Hz

v = n$𝜆$ = 84 $\times$ 4

When wave propagate in air, the velocity

$𝑣=\sqrt{\surd \frac{𝛾𝑝}{𝜌}}$

Here, p = 1.0 $\times$ 10⁵ N/m²

$𝜌$ = 1.2 kg/m³

${𝑣}^2$ = $\frac{𝛾𝑝}{𝜌}$

$𝛾$ = $\frac{{𝑣}^2.𝑝}{𝜌}$

$=\frac{{(84\times 4)}^2\times 1.2}{1.0\times {10}^5}$

$=\frac{135472.2}{{10}^5}$ = 1.354

$\approx$ 1.4

The gas bubble is rising through the liquid of density 175 g/cm³ with constant speed. The acting forces on bubble are



Buoyancy force of liquid = F_b (in upward direction)

Viscous force due to liquid = F_v (in downward direction)

$\therefore$ F_v = F_b

For free falling in a viscous medium

F_v = 6 $𝜋$$𝜂$rv_T

$𝜂$ = viscosity coefficient of the liquid

r = radius of bubble

F_b = mg = v $\times$ $𝜌$ . g

= $\frac{4}{3}$$𝜋$r³$𝜌$_e . g

6$𝜋$$𝜂$rv_t = $\frac{4}{3}$ $𝜋$r³$𝜌$_e . g

$𝜂$ = $\frac{24𝜋.{𝑟}^3{𝜌}_{𝑒}.𝑔}{3\times 6𝜋𝑟{𝑣}_{𝑇}}=\frac{2}{9}.\frac{{𝑟}^2{𝜌}_{𝑒}.𝑔}{{𝑣}_{𝑇}}$

here, D = 2 cm, r = 1 cm = 1 $\times$ 10^$-$2 m

$𝜌$_e = 1.75 g/cm³ = 1.75 $\times$ 10⁺³ kg/m³

g = 10 m/s²

v_T = terminal velocity

= 0.35 $\times$ 10^$-$2 m/s

$𝜂=\frac{2}{9}\times \frac{{\left({10}^{-2}\right)}^2\times 1.75\times {10}^{+3}\times 10}{0.35\times {10}^{-2}}$

$=\frac{2}{9}\times \frac{{10}^{-4}\times 1.75\times {10}^3\times 10}{0.35\times {10}^{-2}}$

$=\frac{2\times 175\times 10\times 10}{9\times 35}\times 10$ poise

$\simeq 1089$ poise

Temperature of water = 0${}^{\circ }$C

Temperature of surrounding atmosphere = $-$ 20${}^{\circ }$C

Density of ice = $𝜌$



Coefficient of thermal conductivity = k

latent heat of melting = L

The rate of heat flow $𝐻=\frac{𝑑𝑄}{𝑑𝑡}=\frac{𝑘𝐴({𝑇}_1-{𝑇}_2)}{𝑥}$

$=\frac{𝑘𝐴\left[\right(0-(-20)]}{𝑥}$

$=\frac{𝑘𝐴\times 20}{𝑥}$

$\frac{𝑑𝑄}{𝑑𝑡}=\frac{𝑑𝑚}{𝑑𝑡}.𝐿$ ($\because$ $𝑚𝐿=𝑄$)

$𝑑𝑚=𝜌𝐴.𝑑𝑥$

$\frac{𝑑𝑄}{𝑑𝑡}=𝜌𝐴.\frac{𝑑𝑥}{𝑑𝑡}𝐿=\frac{𝑘𝐴.20}{𝑥}$

${\int }_{𝑧}^0𝑥.𝑑𝑥=\frac{20𝑘}{𝜌𝐿}{\int }_{𝑡}^0𝑑𝑡\Rightarrow {𝑍}^2=\frac{40𝑘}{𝜌𝐿}𝑡$

So, thickness Z increases as a function of time t according to the above equation.

Let the radius of single drop = t

Radius of small drop = $\frac{2𝑚𝑚}{2}$

= 1 mm

= 1 $\times$ 10^$-$3 m

Surface tension of water = 0.072 N/m

The volume of large drop must be equal volume of all small drops.

$\frac{4}{3}$$𝜋$R³ = 1000$\left(\frac{4}{3}𝜋{𝑟}^3\right)$

$\therefore$ R = (1000)^1/3 . r = 10r

The initial potential energy of drops system U_i = s $\times$ n $\times$ 4$𝜋$r²

= s $\times$ 1000 $\times$ 4$𝜋$r²

= 1000 s . 4$𝜋$r²

When the droplets coalesce, the final potential energy of the system U_f = s . 4$𝜋$R²

= s . 4$𝜋$ $\times$ 100 r²

Energy loss in the formation of large drop

$\Delta$E = U_f $-$ U_i

= s . 4$𝜋$100r² $-$ 1000s . 4$𝜋$r²

= s . $𝜋$r² (400 $-$ 4000)

= $-$ 3600s . $𝜋$r²

= $-$ 3600 $\times$ 0.72 $\times$ 3.14 $\times$ (1 $\times$ 10^$-$3)²

From catenation

= 813.888 $\times$ 10^$-$6

= 8.1388 $\times$ 10^$-$4 J

= 8.146 $\times$ 10^$-$4 J

Breakdown voltage of Zener diode = 5.6 V

Voltage of source = 10 V

Potential difference $\Delta$V = (10 $-$ 5.6) V = 4.4 V

Resistance of circuit = 100 $\Omega$

The current passing through the Zener diode

$𝐼=\frac{\Delta 𝑉}{𝑅}$

$=\frac{4.4}{100}$

$=4.4\times {10}^{-2}$ A

$=4.4\times {10}^{-3}$ A

= 44 mA

Current gain $𝛽$ = 45

We know that, $𝛽=\frac{\Delta {𝐼}_{𝐶}}{\Delta {𝐼}_{𝑏}}$

$\therefore$ $\frac{\Delta {𝐼}_{𝐶}}{\Delta {𝐼}_{𝑏}}=45$ ...... (i)

Potential drop across $1𝑘\Omega$ or $1\times {10}^3\Omega$

Collector resistor is 5 V

$𝑉={𝐼}_{𝐶}𝑅$

${𝐼}_{𝐶}=\frac{𝑉}{𝑅}=\frac{5}{1\times {10}^3}$

$=5\times {10}^{-3}$ A ......(ii)

From Eqs. (i) and (ii),

$\Delta {𝐼}_{𝐵}=\frac{\Delta {𝐼}_{𝐶}}{45}$

$=\frac{5\times {10}^{-3}}{45}=\frac{1}{9}\times {10}^{-3}$

$=0.111\times {10}^{-3}$ A

$=111\times {10}^{-6}$ A = 111 $𝜇$A

Given,

$𝐸=3\widehat{𝑖}+6\widehat{𝑗}+2\widehat{𝑘}$ V/m

$𝐵=2\widehat{𝑖}+3\widehat{𝑗}𝑇$

$𝑣=2\widehat{𝑖}+3\widehat{𝑗}$ m/s

$𝑒=-1.6\times {10}^{-19}$C

The magnitude of electric field

$\left|𝐸\right|=\sqrt{\surd {\left(3\right)}^2+{\left(6\right)}^2+{\left(2\right)}^2}$

$=\sqrt{\surd 9+36+4}$

$=\sqrt{\surd 49}=7$

The force acting on electron due to electric field

$𝐹=𝑞𝐸$

$\left|𝐹\right|=\left|𝑞\right|\left|𝐸\right|$

$=1.6\times {10}^{-19}\times 7𝑁$

$=11.2\times {10}^{-19}𝑁$

$=1.12\times {10}^{-18}𝑁$

Alternate Method

$𝐵\left|\right|𝑣$ Magnetic field is parallel to velocity of electron. $𝑣\left|\right|𝐵$ Magnetic force

${𝐹}_{𝑚}=𝑞(𝑣\times 𝐵)$

$=𝑞𝑣𝐵\sin 𝜃$ ($\because$ $𝜃$ = 0)

$=𝑞𝑣𝐵\sin 0$

$=0$

$\therefore$ Net force $𝐹={𝐹}_{𝑒}+{𝐹}_{𝑚}$

$=(1.12\times {10}^{-18}+0)𝑁$

$=1.12\times {10}^{-18}𝑁$

Given, L₁ = 6 mH = 6 $\times$ 10^$-$3 H

L₂ = 8 mH = 8 $\times$ 10^$-$3 H

If two coils L₁ and L₂ are connected in series, the equivalent self-inductance L for assembly

$𝐿={𝐿}_1+{𝐿}_2+2\sqrt{\surd {𝐿}_1{𝐿}_2}$

$=6\times {10}^{-3}+8\times {10}^{-3}+2\sqrt{\surd 6\times 8\times {10}^{-6}}$

$=14\times {10}^{-3}+2\sqrt{\surd 48}\times {10}^{-3}$

$=14\times {10}^{-3}+2\times 7\times {10}^{-3}$ ($\because$ $\sqrt{\surd 48}=6.92=7$)

$=28\times {10}^{-3}$ H = 28 mH

The time constant of the circuit

$𝜏$ = C. R.

Here, C = 1$𝜇$F = 1 $\times$ 10^$-$6

R = 1 M$\Omega$ = 1 $\times$ 10⁶ $\Omega$

$𝜏$ = 1 $\times$ 10^$-$6 $\times$ 1 $\times$ 10⁶ = 1 s

We know that in one (1) time constant 63% charging is done.

$\therefore$ $\frac{63}{100}\times {𝑞}_{max}$ (q = CV)

$=\frac{63}{100}\times 1\times {10}^{-6}\times 10$

$=\frac{63}{100}\times 1𝜇𝐹\times 10=6.3𝜇𝐶$

$\Rightarrow 𝑉=\frac{𝑞}{𝐶}=\frac{6.3\times {10}^{-6}𝐶}{1\times {10}^{-6}𝐹}=6.3𝑉$

In series combination,

R = R₁ + R₂

Same current will pass in R₁ and R₂. According to question, R₁ is increased by 0.5%.

$\therefore$ Increment in the resistance

= $\frac{0.5}{100}\times 400$

= 2.0 $\Omega$

So to keep the potential unchanged in second resistance the change required will 2.0 $\Omega$ increment.

The electric lines of force are always perpendicular to the equipotential surface

$𝜃$ = 90${}^{\circ }$

The current $𝐼={𝐼}_0{𝑒}^{-𝜆𝑡}$ is of exponential nature.

$𝐼=\frac{𝑑𝑄}{𝑑𝑡}={𝐼}_0{𝑒}^{-𝜆𝑡}$

$𝑑𝑄={𝐼}_0{𝑒}^{-𝜆𝑡}𝑑𝑡$

$\underset{0}{\overset{𝑄}{\int }}𝑑𝑄={𝐼}_0\underset{𝑡=0}{\overset{𝑡=\infty }{\int }}{𝑒}^{-𝜆𝑡}.𝑑𝑡$

$𝑄={𝐼}_0\times \frac{1}{𝜆}=\frac{{𝐼}_0}{𝜆}$

For Fraunhoffer diffraction to occur, both source and screen should be at infinity, it is essential condition for diffraction pattern.

Information in the statement of the questions is insufficient to calculate heat required to raise the temperature from 100${}^{\circ }$C to 1000${}^{\circ }$C.

(i) When the body of mass m lifted up the forces acting on the body



${𝑇}_1-𝑚𝑔=\frac{𝑚𝑔}{2}$

${𝑇}_1=𝑚𝑔+\frac{𝑚𝑔}{2}=\frac{3𝑚𝑔}{2}$

(ii) When the body lowered the acting forces



$(𝑚𝑔-{𝑇}_2)=\frac{𝑚𝑔}{2}$

${𝑇}_2=𝑚𝑔-\frac{𝑚𝑔}{2}=\frac{𝑚𝑔}{2}$

On dividing Eq. (i) from Eq (ii) we get

$\therefore$ $\frac{{𝑇}_1}{{𝑇}_2}=\frac{3𝑚𝑔/2}{𝑚𝑔/2}=\frac{3}{1}$

= 3 : 1

Let when link AB is removed, x is an equivalent of remaining part of cube without link.

AB and rest part will be in parallel, so according to question,

$\frac{7}{12}=\frac{{𝑅}_1{𝑅}_2}{{𝑅}_1+𝑅}=\frac{1\left(𝑥\right)}{1+𝑥}=\frac{𝑥}{1+𝑥}$

$\frac{7}{12}=\frac{𝑥}{1+𝑥}$

$12𝑥=7+7𝑥$

$12𝑥-7𝑥=7$

$5𝑥=7$

$𝑥=\frac{7}{5}\Omega$

The necessary centripetal force, to move in a circular path is provided by the Coulomb's force.

$\frac{{𝑚}_1{𝑣}^2}{𝑟}=\frac{1}{4𝜋{𝜀}_0}.\frac{{𝑞}_1{𝑞}_2}{{𝑟}^2}$

$𝑣=\sqrt{\surd \frac{1}{4𝜋{𝜀}_0}.\frac{{𝑞}_1{𝑞}_2}{𝑟{𝑚}_1}}$ .... (i)

We know, $𝑣=𝑟𝜔$

$𝜔=\frac{2𝜋}{𝑇}$

$𝑇=\frac{2𝜋}{𝜔}$

$=\frac{2𝜋.𝑟}{𝑣}$ .... (ii)

Substitution the value of v from, Eq. (i) we get

$𝑇=\frac{2𝜋𝑟}{\sqrt{\surd \frac{{𝑞}_1{𝑞}_2}{4𝜋{𝜀}_0𝑟.{𝑚}_1}}}$

$𝑇=\sqrt{\surd \frac{4{𝜋}^2{𝑟}^2\times 4𝜋{𝜀}_0.𝑟.{𝑚}_1}{{𝑞}_1{𝑞}_2}}$

$𝑇=\sqrt{\surd \frac{16{𝜋}^3{𝜀}_0{𝑟}^3.{𝑚}_1}{{𝑞}_1{𝑞}_2}}$

Both the charges are positive or both negative. Since, answers have the product q₁q₂ inside square root. Therefore, circular motion is not possible.

The statement of question is wrong. Either q₁ or q₂ should be negative.

The intensity of emitted electrons is

$𝐼\propto \frac{1}{{𝑟}^2}$

${𝐼}_1\propto \frac{1}{{𝑑}^2}$

${𝐼}_2\propto \frac{1}{{(𝑑/2)}^2}=4.\frac{1}{{𝑑}^2}$

and I $\propto$ Number of photons per second $\propto$ N

$\therefore$ $𝑁\propto \frac{1}{{𝑟}^2}$

$\therefore$ Number of emitted electrons will become 4 times. From Einstein's photo-electric equation

$𝐾{𝐸}_{max}=ℎ𝑣-𝜙$

where, $𝜙$ = work function of surface

v = frequency of incident photon

Since, v remains unchanged, thus KE_max as well as stopping potential remains unchanged.

$\therefore$ $𝐾{𝐸}_{max}=𝑒{𝑉}_{𝑠}\propto 𝑣$

where, V_s = stopping potential.