NEET-UG 2015

$$\upsilon$$_c = $\left[{\eta }^x{\rho }^y{r}^z\right]$

Put dimensions of various quantities,

[M⁰LT^-1] = [ML^-1T^-1]^x [ML^-3T⁰]^y [M⁰LT⁰]^z
                = [M^{x + y} L^{- x - 3y + z} T^{- x}]

Equating power both sides, we get

x + y = 0;
- x - 3y + z = 1;
- x = - 1

On solving, we get

x = 1, y = -1, z = -1

Here, $\overrightarrow{R}=4\sin \left(2\pi t\right)\hat{i}+4\cos \left(2\pi t\right)\hat{j}$

The velocity of the particle is
$\overrightarrow{v}=\frac{d\overrightarrow{R}}{dt}=\frac{d}{dt}\left[4\sin \left(2\pi t\right)\hat{i}+4\cos \left(2\pi t\right)\hat{j}\right]$
$=8\pi cos\left(2\pi t\right)\hat{i}-8\pi sin\left(2\pi t\right)\hat{j}$

Its magnitude is

$\left|\overrightarrow{v}\right|=\sqrt{{\left(8\pi \cos \left(2\pi t\right)\right)}^2+{\left(-8\pi \sin \left(2\pi t\right)\right)}^2}$

$=\sqrt{64{\pi }^2{\cos }^2\left(2\pi t\right)+64{\pi }^2{\sin }^2\left(2\pi t\right)}$

$=\sqrt{64{\pi }^2\left[{\cos }^2\left(2\pi t\right)+{\sin }^2\left(2\pi t\right)\right]}$

$=\sqrt{64{\pi }^2}$     (as sin²$\theta$ + cos²$\theta$ = 1)

= 8$\pi$ m/s

Two vectors $\overline{A}$ and $\overline{B}$ are orthogonal to each other, if their scalar product is zero i.e. $\overline{A}$.$\overline{B}$ = 0.

Here, $\overline{A}=\cos \omega t\hat{i}+\sin \omega t\hat{j}$

and $\overline{B}=\cos \frac{\omega t}{2}\hat{i}+\sin \frac{\omega t}{2}\hat{j}$

$\therefore \overline{A}.\overline{B}=\left(\cos \omega t\hat{i}+\sin \omega t\hat{j}\right)\left(\cos \frac{\omega t}{2}\hat{i}+\sin \frac{\omega t}{2}\hat{j}\right)$

= $\cos \omega t\cos \frac{\omega t}{2}+\sin \omega t\sin \frac{\omega t}{2}$
    $\left(\because \hat{i}.\hat{i}=\hat{j}.\hat{j}=1and\hat{i}.\hat{j}=\hat{j}.\hat{i}=0\right)$

= $\cos \left(\omega t-\frac{\omega t}{2}\right)$
    ($\because \cos (A-B)=cosAcosB+sinAsinB$)

But $\overline{A}.\overline{B}=0$ (as $\overline{A}$ and $\overline{B}$ are orthogonal to each other)

$\therefore \cos \left(\omega t-\frac{\omega t}{2}\right)=0$

$\cos \left(\omega t-\frac{\omega t}{2}\right)=\cos \frac{\pi }{2}or\omega t-\frac{\omega t}{2}=\frac{\pi }{2}$

$\frac{\omega t}{2}=\frac{\pi }{2}ort=\frac{\pi }{\omega }$

According to question, two stones experience same centripetal force  i.e. F_C1 = F_C2

or $\frac{m{v}_1^2}{r}=\frac{2m{v}_2^2}{\left(r/2\right)}\Rightarrow V_1^2=4V_2^2$

So, V₁ = 2V₂ i.e., n = 2

Coefficient of static friction,
${\mu }_s=\tan 30^{\circ }=\frac{1}{\sqrt{3}}=0.577\cong 0.6$

$S=ut+\frac{1}{2}a{t}^2$

$4=\frac{1}{2}a{\left(4\right)}^2\Rightarrow a=\frac{1}{2}=0.5$
[$\because$ s = 4m and t = 4s given]

a = gsin$\theta$ – ${\mu }_k(g)\cos \theta$

$\Rightarrow {\mu }_k=\frac{0.9}{\sqrt{3}}=0.5$

When ball collides with the ground it loses its 50% of energy

$\frac{K.E_f}{K.E_i}=\frac{1}{2}\Rightarrow \frac{\frac{1}{2}m{V}_f^2}{\frac{1}{2}m{V}_i^2}=\frac{1}{2}$

$\Rightarrow \frac{V_f}{V_i}=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{\sqrt{2gh}}{\sqrt{V_0^2+2gh}}=\frac{1}{\sqrt{2}}$

or $4gh=V_0^2+2gh$

$\therefore$ V₀ = 20 ms^–1

For collision ${\overrightarrow{V}}_{B/A}$ should be along
$\overline{B\to A}\left({\overrightarrow{r}}_{A/B}\right)$

So, $\frac{\overrightarrow{V_2}-\overrightarrow{V_1}}{\left|\overrightarrow{V_2}-\overrightarrow{V_1}\right|}=\frac{\overrightarrow{r_1}-\overrightarrow{r_2}}{|\overrightarrow{r_1}-\overrightarrow{r_2}|}$

The situation is shown in the figure. Let v' be speed of second block after the collision. As the collision is elastic, so kinetic energy is conserved.
According to conservation of kinetic energy,

$\frac{1}{2}M{v}^2+0=\frac{1}{2}M{\left(\frac{v}{3}\right)}^2+\frac{1}{2}Mv{}^{\prime 2}$

$v^2=\frac{v^2}{9}+v{}^{\prime 2}$

$\Rightarrow v{}^{\prime 2}=v^2-\frac{v^2}{9}=\frac{9v^2-v^2}{9}=\frac{8}{9}{v}^2$

$v^{\prime }=\sqrt{\frac{8}{9}{v}^2}=\frac{\sqrt{8}}{3}v=\frac{2\sqrt{2}}{3}v$

Here, Volume of blood pumped by man’s heart,
V = 5 litres = 5 × 10^–3 m³ ($\because$ 1 litre = 10–3 m³)

Time in which this volume of blood pumps, t = 1 min = 60 s
Pressure at which the blood pumps,

P = 150 mm of Hg = 0.15 m of Hg
= (0.15 m)(13.6 × 10³ kg/m³)(10 m/s²) ($\because$ $P=h\rho g$)

= 20.4 × 103 N/m²

$\therefore$ Power of the heart = $\frac{PV}{t}$

$=\frac{\left(20.4\times {10}^{3}N/m^2\right)\left(5\times {10}^{-3}{m}^3\right)}{60s}=1.70W$

Here, Speed of the automobile,

$v=54km{h}^{-1}=54\times \frac{5}{18}m{s}^{-1}=15m{s}^{-1}$

Radius of the wheel of the automobile, R = 0.45 m
Moment of inertia of the wheel about its axis of rotation, I = 3 kg m²

Time in which the vehicle brought to rest, t = 15 s
The initial angular speed of the wheel is

${\omega }_i=\frac{v}{R}=\frac{15m{s}^{-1}}{0.45m}=\frac{1500}{45}rad{s}^{-1}=\frac{100}{3}rad{s}^{-1}$

and its final angular speed is ${\omega }_f$ = 0 (as the vehicle comes to rest)

$\therefore$ The angular retardation of the wheel is

$\alpha =\frac{{\omega }_f-{\omega }_i}{t}=\frac{0-\frac{100}{3}}{15s}=-\frac{100}{45}rad{s}^{-1}$

The magnitude of required torque is

$\tau =I\left|\alpha \right|=\left(3kg{m}^2\right)\left(\frac{100}{45}rad{s}^{-2}\right)$

$=\frac{20}{3}kg{m}^2{s}^{-2}=6.66kg{m}^2{s}^{-2}$

$K=\frac{1}{2}\left(m_1{x}^2\right){\omega }_0^2+\frac{1}{2}{m}_2{\left(L-x\right)}^2{\omega }_0^2$

$dK/dx=0$

$m_{1}x=m_2\left(L-x\right)$

$m_{1}x=m_{2}L-m_{2}x$

$\Rightarrow x=\frac{m_{2}L}{m_1+m_2}$

From Newton's second law for rotational motion,

$\overrightarrow{\tau }=\frac{\overrightarrow{d}L}{dt}$, if $\overrightarrow{L}$ = constant then $\overrightarrow{\tau }=0$

So, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=0$

$\left(2\hat{i}-6\hat{j}-12\hat{k}\right)\times \left(\alpha \hat{i}+3\hat{j}+6\hat{k}\right)=0$

Solving we get $\alpha$ = –1

The gravitational force on the satellite will be aiming towards the centre of the earth so acceleration of the satellite will also be aiming towards the centre of the earth.

The orbital speed of the satellite is

$v_o=R\sqrt{\frac{g}{\left(R+h\right)}}$

where R is the earth’s radius, g is the acceleration due to gravity on earth’s surface and h is the height above the surface of earth.

Here, R = 6.38 × 10⁶m, g = 9.8 m s^–2 and h = 0.25 × 10⁶ m

$v_o=\left(6.38\times {10}^{6}m\right)\sqrt{\frac{9.8m{s}^{-2}}{\left(6.38\times {10}^{6}m+0.25\times {10}^{6}m\right)}}$

= 7.76 × 10³ m s^–1 = 7.76 km s^–1

Young's modulus $Y=\frac{W}{A}.\frac{l}{\Delta l}$

$\frac{W_1}{Y_1}=\frac{W_2}{Y_2}$
[$\because A,l,\Delta l$ same for both brass and steel]

$\frac{W_1}{W_2}=\frac{Y_1}{Y_2}=2$
  $\left[\frac{Y_{steel}}{Y_{brass}}=2given\right]$

Let the speed of the ejection of the liquid through the holes be v. Then according to the equation of continuity,

$\pi R^{2}V=n\pi r^{2}v$

$\Rightarrow v=\frac{\pi R^{2}V}{n\pi r^2}=\frac{V{R}^2}{n{r}^2}$

Water will not overflow but will change its radius of curvature.

Let r₀ and r_T be densities of glycerin at 0°C and T°C respectively. Then,

${\rho }_T={\rho }_0\left(1-\gamma \Delta T\right)$

where $\gamma$ is the coefficient of volume expansion of glycerine and $\Delta T$ is rise in temperature.

$\frac{{\rho }_T}{{\rho }_0}=1-\gamma \Delta T\Rightarrow \gamma \Delta T=1-\frac{{\rho }_T}{{\rho }_0}$

Thus, $\frac{{\rho }_0-{\rho }_T}{{\rho }_0}=\gamma \Delta T$

Here, $\gamma =5\times 10^{-4}{K}^{-1}$ and $\Delta T=40^{o}C=40K$

$\therefore$ The fractional change in the density of glycerin

$=\frac{{\rho }_0-{\rho }_T}{{\rho }_0}=\gamma \Delta T=\left(5\times {10}^{-4}{K}^{-1}\right)\left(40K\right)=0.020$

The P-V diagram of an ideal gas compressed from its initial volume $V_0$ to $\frac{V_0}{2}$ by several processes is shown in the figure.



Work done on the gas = Area under P–V curve

As area under the P–V curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.

Coefficient of performance,

$Cop=\frac{T_2}{T_1-T_2}$

$5=\frac{273-20}{T_1-\left(273-20\right)}=\frac{253}{T_1-253}$

5T₁ – (5 × 253) = 253

5T₁ = 253 + (5 × 253) = 1518

$\therefore$ $T_1=\frac{1518}{5}=303.6$

$\Rightarrow$ T₁ = 303.6 – 273 = 30.6 $\cong$ 31°C

From PV = nRT

$P_A=\frac{{\rho }_A{M}_A}{RT}$ and $P_B=\frac{{\rho }_B{M}_B}{RT}$

From question,

$\frac{P_A}{P_B}=\frac{{\rho }_A}{{\rho }_B}\frac{M_A}{M_B}=2\frac{M_A}{M_B}=\frac{3}{2}$

$\therefore$ $\frac{M_A}{M_B}=\frac{3}{4}$

As, we know, in SHM
Maximum acceleration of the particle, $\alpha$ = A${\omega }^2$
Maximum velocity, $\beta$ = A$\omega$

$\Rightarrow \omega =\frac{\alpha }{\beta }$

$\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi \beta }{\alpha }$   $\left[\because \omega =\frac{2\pi }{T}\right]$

In a stretched string all multiples of frequencies can be obtained i.e., if fundamental frequency is n then higher frequencies will be 2n, 3n, 4n ...



So, the difference between any two successive frequencies will be 'n'

According to question, n = 420 – 315 = 105Hz

So the lowest frequency of the string is 105 Hz.

Fundamental frequency of closed organ pipe

$V_c=\frac{V}{4l_c}$

Fundamental frequency of open organ pipe

$V_0=\frac{V}{2l_0}$

Second overtone frequency of open organ pipe

$=\frac{3V}{2l_0}$

$\frac{V}{4l_c}=\frac{3V}{2l_0}$

$\Rightarrow l_0=6l_c=6\times 20=120$ cm

Here, original frequency of sound, f₀ = 100 Hz
Speed of source V_s = 19.4 cos 60° = 9.7



From Doppler's formula

$f^1=f_0\left(\frac{V-V_0}{V-V_S}\right)$

$f^1=100\left(\frac{V-0}{V-\left(+9.7\right)}\right)$

$f^1=100\frac{V}{V\left(1-\frac{9.7}{V}\right)}=\frac{100}{\left(1-\frac{9.7}{330}\right)}$

= 103Hz

Apparent frequency f¹ = 103 Hz

Molar mass of the gas = 4g/mol
Speed of sound

$V=\sqrt{\frac{\gamma RT}{m}}\Rightarrow 952=\sqrt{\frac{\gamma \times 3.3\times 273}{4\times {10}^{-3}}}$

$\Rightarrow \gamma =1.6=\frac{16}{10}=\frac{8}{5}$

Also, $\gamma =\frac{C_P}{C_V}=\frac{8}{5}$

$\Rightarrow C_P=\frac{8\times 5}{5}=8J{K}^{-1}mo{l}^{-1}$
[C_V = 5.0 JK^–1 given]

Potential in a region V = 6xy – y + 2yz

As we know the relation between electric
potential and electric field is $\overrightarrow{E}=\frac{-dV}{dx}$

$\overrightarrow{E}=\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)$

$\overrightarrow{E}=\left[\left(6y\hat{i}+\left(6x-1+2z\right)\hat{j}+\left(2y\right)\hat{k}\right)\right]$

${\overrightarrow{E}}_{\left(1,1,0\right)}=-\left(6\hat{i}+5\hat{j}+2\hat{k}\right)$

In figure, two metal wires of identical dimension are connected in series


$R_{eq}=\frac{1}{{\sigma }_{1}A}+\frac{1}{{\sigma }_{2}A}=\frac{l_{eq}}{{\sigma }_{eq}{A}_{eq}}$

$\frac{2l}{{\sigma }_{eq}A}=\frac{1}{A}\left(\frac{{\sigma }_1+{\sigma }_2}{{\sigma }_1{\sigma }_2}\right)$

$\therefore {\sigma }_{eq}=\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$

From circuit diagram


Resistance of ammeter
= $\frac{480\times 20}{480+20}=19.2\Omega$

Total resistance R = 40.8 + 19.2 = 60$\Omega$

Reading in the ammeter $i=\frac{V}{R}$

$=\frac{30}{40.8+19.2}=0.5A$

The current through the potentiometer wire is
$I=\frac{E_0}{\left(r+r_1\right)}$

and the potential difference across the wire is

$V=Ir=\frac{E_{0}r}{\left(r+r_1\right)}$

The potential gradient along the potentiometer wire is

$k=\frac{V}{L}=\frac{E_{0}r}{\left(r+r_1\right)L}$

As the unknown e.m.f. E is balanced against length l of the potentiometer wire,

$\therefore$ $E=kl=\frac{E_{0}r}{\left(r+r_1\right)}\frac{l}{L}$

Force of attraction between the plates, F = qE

$=q\times \frac{\sigma }{2{\in }_0}=q\frac{q}{2A{\in }_0}$

$=\frac{q^2}{2\left(\frac{{\in }_{0}A}{d}\right)\times d}=\frac{C^2{V}^2}{2cd}=\frac{C{V}^2}{2d}$

Here, $C=\frac{{\in }_{0}A}{d}$, $q=CV$, A = area

As we know, F = qvB = $\frac{m{v}^2}{R}$

$\therefore$ $R=\frac{mv}{qB}=\frac{\sqrt{2m\left(kE\right)}}{qB}$

Since R is same so, $KE\propto \frac{q^2}{m}$

Therefore KE of $\alpha$ particle

$=\frac{q^2}{m}=\frac{{\left(2\right)}^2}{4}=1MeV$

The required torque is $\tau$ = NIABsin$\theta$

= (50)(2 A)(0.012 m²)(0.2 Wb/m²) sin60°

= 0.20 N m

When e^– is coming towards the loop magnetic flux of one type increased and when going away same magnetic flux decreased so induced current opposite to each other.

For series R – C circuit, capacitive reactance,

Z= $\sqrt{R^2+{\left(\frac{1}{C\omega }\right)}^2}$

Current through resistor, i = Current in the circuit

= $\frac{V}{Z_c}=\frac{V}{\sqrt{R^2+{\left(\frac{1}{C\omega }\right)}^2}}$

Voltage across capacitor, V = iX_C

= $\frac{V}{\sqrt{R^2+{\left(\frac{1}{C\omega }\right)}^2}}\times \frac{1}{C\omega }$

= $\frac{V}{\sqrt{R^2{\omega }^2{C}^2+1}}$

When mica is introduced capacitance will increase, hence voltage across capacitor gets decrease.

Energy of x-ray is (100 eV to 100 keV).

Hence energy of the order of 15 keV belongs to X-rays.

Magnification by eye piece

m = $\frac{f}{f+u}$

$\Rightarrow$ $-\frac{I}{L}=\frac{f_e}{f_e+\left[-\left(f_0+f_e\right)\right]}$ = - $\frac{f_e}{f_0}$

$\Rightarrow$ $\frac{I}{L}=\frac{f_e}{f_0}$

$\therefore$ Magnification, M = $\frac{f_0}{f_e}$ = $\frac{L}{I}$

for no emergence i $>$ C.

sin i $>$ sin C

$\Rightarrow$ $\frac{1}{\sqrt{2}}>\frac{1}{\mu }$

$\Rightarrow$ $\mu >\sqrt{2}$ = 1.414

As $\mu$_red (= 1.39) < $\mu$(= 1.414) while $\mu$_green( = 1.44) and $\mu$_blue(= 1.47) > $\mu$ (= 1.414), so only red colour will be transmitted through face AC while green and blue colours will suffer total internal reflection.

So the prism will separate red colour from the green and blue colours.

Given, The ratio of slits width = $\frac{1}{25}$

$\therefore$ $\frac{I_1}{I_2}=\frac{25}{1}$

I $\propto$ A²

$\Rightarrow$ $\frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}$ = $\frac{25}{1}$

$\Rightarrow$ $\frac{A_1}{A_2}=\frac{5}{1}$

$\frac{A_{\max }}{A_{\min }}=\frac{A_1+A_2}{A_1-A_2}$ = $\frac{5+1}{5-1}$ = $\frac{6}{4}$ = $\frac{3}{2}$

$\therefore$ $\frac{I_{\max }}{I_{\min }}=\frac{A_{\max }^2}{A_{\min }^2}$ = ${\left(\frac{3}{2}\right)}^2$ = $\frac{9}{4}$

For first minima at P

AP – BP = $\lambda$

AP – MP = $\frac{\lambda }{2}$

So phase difference, $\varphi$ = $\frac{2\pi }{\lambda }\times \frac{\lambda }{2}$ = $\pi$ radian

In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity.

U $\to$ Th + He

KE_Th = $\frac{P^2}{2m_{Th}}$

KE_He = $\frac{P^2}{2m_{He}}$

since m_He is less so KE_He will be more.

For the longest wavelength in the Lyman series,

n₁ = 1 and n₂ = 2

$\frac{1}{{\lambda }_L}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ = $\frac{3R}{4}$

For the longest wavelength in the Balmer series,

n₁ = 2 and n₂ = 3

$\frac{1}{{\lambda }_B}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ = $\frac{5R}{36}$

$\therefore$ $\frac{{\lambda }_L}{{\lambda }_B}=\frac{\frac{4}{3R}}{\frac{36}{5R}}$ = $\frac{5}{27}$

KE_max = $\frac{hc}{\lambda }$ - $\varphi$

$\Rightarrow$ KE_max = $\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{5\times {10}^{-7}\times 1.6\times {10}^{-19}}$ - 2.28

$\Rightarrow$ KE_max = 2.48 – 2.28 = 0.2 eV

$\lambda$_min = $\frac{h}{\sqrt{2mK{E}_{\max }}}$

= $\frac{\frac{20}{3}\times {10}^{34}}{\sqrt{2\times 9\times {10}^{-31}\times 0.2\times 1.6\times {10}^{-19}}}$

= 2.8 $\times$ 10^$-$9 m

$\therefore$ $\lambda$ $\ge$ 2.8 $\times$ 10^$-$9 m

According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is

$K_{{\max }_1}$ = $\frac{hc}{\lambda }$ - $\varphi$₀

and that in the second case is

$K_{{\max }_2}$ = $\frac{hc}{\frac{\lambda }{2}}$ - $\varphi$₀

Given $K_{{\max }_2}$ = 3$K_{{\max }_1}$

$\therefore$ $\frac{2hc}{\lambda }$ - $\varphi$₀ = 3$\left(\frac{hc}{\lambda }-{\varphi }_0\right)$

$\Rightarrow$ $\frac{2hc}{\lambda }$ - $\varphi$₀ = $\frac{3hc}{\lambda }-3{\varphi }_0$

$\Rightarrow$ 2$\varphi$₀ = $\frac{hc}{\lambda }$

$\Rightarrow$ $\varphi$₀ = $\frac{hc}{2\lambda }$

Given, V_i = 2cos(15t + $\frac{\pi }{3}$)

and Voltage gain A_V = 150

For CE transistor phase difference between input and output signal is $\pi$ = 180^o

Using formula, A_V = $\frac{V_0}{V_i}$

$\Rightarrow$ V₀ = A_V $\times$ V_i

= 150$\times$2cos(15t + $\frac{\pi }{3}$ + $\pi$)

= 300cos$\left(15t+\frac{4\pi }{3}\right)$

Potential difference across resistance R

= 3.5 – 0.5 = 3.0 V

Current in circuit, I = V/R = 3/100 = 30 mA

50 ml of 16.9% solution of AgNO₃

$\left(\frac{16.9}{100}\times 50\right)$ = 8.45 g of AgNO₃

n_mole = $\frac{8.45g}{(107.8+14+16\times 3)g/mol}$

= $\left(\frac{8.45g}{169.8g/mol}\right)=0.0497moles$

50ml of 5.8% solution of NaCl contain

NaCl = $\left(\frac{5.8}{100}\times 50\right)=2.9g$

n_NaCl = $\frac{2.9g}{(23+35.5)g/mol}$

= 0.0495 moles

AgNO3	+	NaCl	$$\Rightarrow$$	AgCl	+	Na$$\oplus$$	+	Cl$$\ominus$$
1 mole		1 mole		1 mole
0.049 mole		0.049 mole		0.049 mole of AgCl

n = $\frac{w}{M}$

$\Rightarrow$ w = (n_AgCl) $\times$ Molecular mass

= (0.049) $\times$ (107.8 + 35.5) = 7.02 g

We know mass of 1 mol (6.022 $\times$ 10²³) atoms of carbon = 12 g

If Avogadro number is changed to 6.022 $\times$ 10²⁰ then mass of 1 mol of cabon is

= $\frac{12\times 6.022\times {10}^{20}}{6.022\times {10}^{23}}$ = $12\times 10^{-3}g$

$\therefore$ It would change the mass of one mole of carbon.

Number of moles of water in 1000 g = $\frac{1000}{18}$ = 55.56 mol

Thus, mole fraction of solute = $\frac{1}{1+55.56}$ = 0.0177