JEE Main 2026 24th January Evening Shift Question Paper with Solutions

We need to find the intensity on the screen in front of one of the slits in Young's double slit experiment. The intensity from each slit is $I_0$, the slit separation is d = 2 mm = 2 × 10⁻³ m, the screen distance is D = 10 m, and the wavelength is $\lambda = 6000 \text{ Å} = 6 \times 10^{-7}$ m.

The fringe width is given by $\beta = \frac{\lambda D}{d} = \frac{6 \times 10^{-7} \times 10}{2 \times 10^{-3}} = 3 \times 10^{-3} \text{ m} = 3 \text{ mm}$.

The point in front of one slit lies at y = d/2 = 1 mm from the central maximum, so the path difference is $\Delta = \frac{yd}{D} = \frac{1 \times 10^{-3} \times 2 \times 10^{-3}}{10} = 2 \times 10^{-7}$ m.

The corresponding phase difference is $\phi = \frac{2\pi}{\lambda} \times \Delta = \frac{2\pi}{6 \times 10^{-7}} \times 2 \times 10^{-7} = \frac{2\pi}{3}$.

The resultant intensity is $I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0}\cos\phi = 2I_0 + 2I_0\cos\frac{2\pi}{3},$ which simplifies to $= 2I_0 + 2I_0\left(-\frac{1}{2}\right) = 2I_0 - I_0 = I_0.$

Therefore, the intensity is Option 3: $I_0$.

We need to find the wavelength of the first light given photoelectric stopping potentials. For light of wavelength $\lambda$ the stopping potential is $V_1 = 3.2$ V, and for light of wavelength $2\lambda$ the stopping potential is $V_2 = 0.7$ V.

According to Einstein's photoelectric equation, $eV = \frac{hc}{\lambda} - \phi.$ Applying this to the first light gives $eV_1 = \frac{hc}{\lambda} - \phi \quad \cdots (1)$ and to the second light gives $eV_2 = \frac{hc}{2\lambda} - \phi \quad \cdots (2).$

Subtracting equation (2) from (1) yields $e(V_1 - V_2) = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda},$ which leads to $\lambda = \frac{hc}{2e(V_1 - V_2)}.$

Substituting the numerical values, we get $\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2 \times 1.6 \times 10^{-19} \times (3.2 - 0.7)}$ $= \frac{19.89 \times 10^{-26}}{2 \times 1.6 \times 10^{-19} \times 2.5}$ $= \frac{19.89 \times 10^{-26}}{8 \times 10^{-19}} = 2.486 \times 10^{-7} \text{ m} \approx 2.5 \times 10^{-7} \text{ m}.$

Therefore, the wavelength is Option 2: $2.5 \times 10^{-7}$ m.

We need to find the height of the submerged part of a floating cubical block.

The density of the block is $\rho_b = 600$ kg/m³, and the density of the liquid is $\rho_l = 900$ kg/m³. The block has height H = 8.0 cm.

By the principle of floatation, the weight of the block equals the weight of the liquid displaced, so

$\rho_b \times V_{block} \times g = \rho_l \times V_{submerged} \times g$

Because the block is cubical with base area A, its total volume is A × H and its submerged volume is A × h, giving

$\rho_b \times A \times H = \rho_l \times A \times h$

Solving for h yields

$h = \frac{\rho_b}{\rho_l} \times H = \frac{600}{900} \times 8.0 = \frac{2}{3} \times 8.0 = 5.33 \text{ cm} \approx 5.3 \text{ cm}$

Therefore, the height of the submerged part is Option 4: 5.3 cm.

The stability of a nucleus is measured by its total binding energy: larger binding energy ⇒ greater stability.
Let the binding energies of ${}_{2}He^{3},\; {}_{2}He^{4}$ and ${}_{2}He^{5}$ be $B_3,\; B_4$ and $B_5$ (in MeV).

Write the given reactions in the form
${}_{2}He^{3} + {}_{0}n^{1} \;\longrightarrow\; {}_{2}He^{4} + 20\;{\rm MeV}$
${}_{2}He^{4} + {}_{0}n^{1} \;\longrightarrow\; {}_{2}He^{5} - 0.9\;{\rm MeV}$

For any nuclear reaction,
energy released $=\;(\text{binding energy of products})-(\text{binding energy of reactants})$.

Case 1:
Products’ binding energy $=\;B_4$, reactants’ binding energy $=\;B_3 + 0$ (a free neutron has zero binding energy).
Hence
$B_4 - (B_3 + 0) = 20$ $-(1)$
$\Rightarrow\; B_4 = B_3 + 20$

Case 2:
Products’ binding energy $=\;B_5$, reactants’ binding energy $=\;B_4 + 0$.
Here the reaction absorbs 0.9 MeV, so energy released = $-0.9$ MeV:
$B_5 - (B_4 + 0) = -0.9$ $-(2)$
$\Rightarrow\; B_5 = B_4 - 0.9$

Substitute $B_4$ from $(1)$ into $(2)$:
$B_5 = (B_3 + 20) - 0.9 = B_3 + 19.1$

Thus
$B_4 = B_3 + 20$
$B_5 = B_3 + 19.1$
Since $20 \gt 19.1 \gt 0$, we get the order
$B_4 \gt B_5 \gt B_3$.

Greater binding energy means higher stability, so
$X_4 \gt X_5 \gt X_3$.

Therefore the correct option is Option A.

$= A(\overline{A} + \overline{B})$
$= A(\overline{B})$

Hence, (A = 1) and (B = 0) for (Y = 1). Otherwise, (Y = 0).
Hence, option (B) is correct.

Label the six vertices of the regular hexagon in order as $A,\,B,\,C,\,D,\,E,\,F$ and let the centre be $O$. Each side and each spoke (vertex-centre wire) has resistance $r$.

The current enters at vertex $A$ and leaves at the opposite vertex $D$. Because the geometrical arrangement is symmetric about the line $AD$, the following potentials will be equal:   • $V_B = V_F = x$   (call this common potential $x$)
  • $V_C = V_E = y$   (call this $y$)
  • $V_O = z$   (centre potential)

Let the potential at $A$ be $V_A = V$ (unknown for now) and at $D$ be $V_D = 0$. We shall write Kirchhoff’s Current Law (KCL) at the three distinct internal potentials $x,\,y,\,z$. (All currents are taken as leaving the node, and each resistance is $r$.)

Case 1: Node $B$ (or $F$) at potential $x$
Connections: to $A$, $C$, $O$. $\frac{x - V}{r} + \frac{x - y}{r} + \frac{x - z}{r} = 0$ Multiplying by $r$ gives: $3x - V - y - z = 0 \quad -(1)$

Case 2: Node $C$ (or $E$) at potential $y$
Connections: to $B$, $D$, $O$. $\frac{y - x}{r} + \frac{y - 0}{r} + \frac{y - z}{r} = 0$ Multiplying by $r$ gives: $3y - x - z = 0 \quad -(2)$

Case 3: Centre $O$ at potential $z$
Connections: to all six vertices. $\frac{z - V}{r} + \frac{z - x}{r} + \frac{z - y}{r} + \frac{z - 0}{r} + \frac{z - y}{r} + \frac{z - x}{r} = 0$ Grouping like terms and multiplying by $r$: $6z - V - 2x - 2y = 0 \quad -(3)$

Solving the simultaneous equations

From $-(1)$:  $x = \frac{V + y + z}{3} \quad -(1a)$

From $-(2)$:  $y = \frac{x + z}{3} \quad -(2a)$

Substitute $x$ from $(1a)$ into $(2a)$:
$y = \frac{\dfrac{V + y + z}{3} + z}{3} = \frac{V + y + 4z}{9}$
$\Rightarrow 9y = V + y + 4z \Longrightarrow 8y = V + 4z$
$\therefore \; y = \frac{V + 4z}{8} \quad -(4)$

Put $y$ from $(4)$ into $(1a)$:
$x = \frac{V + \dfrac{V + 4z}{8} + z}{3} = \frac{9V + 12z}{24} = \frac{3V + 4z}{8} \quad -(5)$

Now use equation $-(3)$:
$6z - V - 2x - 2y = 0$
Insert $x$ and $y$ from $-(5)$ and $-(4)$:

$6z - V - 2\!\left(\frac{3V + 4z}{8}\right) - 2\!\left(\frac{V + 4z}{8}\right)=0$
$\Longrightarrow 6z - V - \frac{3V + 4z}{4} - \frac{V + 4z}{4}=0$
$\Longrightarrow 6z - V - (V + 2z)=0$
$\Longrightarrow 4z - 2V=0 \;\; \Rightarrow\;\; z = \frac{V}{2}$

Back-substitute $z$:

$y = \frac{V + 4\!\left(\dfrac{V}{2}\right)}{8} = \frac{3V}{8}, \qquad x = \frac{3V + 4\!\left(\dfrac{V}{2}\right)}{8} = \frac{5V}{8}$

Total current entering at $A$
Three resistors leave $A$: to $B$, $O$, $F$. Hence $I = \frac{V - x}{r} + \frac{V - z}{r} + \frac{V - x}{r} = \frac{2(V - x) + (V - z)}{r}$

Substitute $x = \dfrac{5V}{8}, \; z = \dfrac{V}{2}$:
$I = \frac{2\!\left(V - \dfrac{5V}{8}\right) + \left(V - \dfrac{V}{2}\right)}{r} = \frac{2\!\left(\dfrac{3V}{8}\right) + \dfrac{V}{2}}{r} = \frac{\dfrac{6V}{8} + \dfrac{4V}{8}}{r} = \frac{\dfrac{10V}{8}}{r} = \frac{5V}{4r}$

Equivalent resistance
$R_{\text{eq}} = \frac{V}{I} = \frac{V}{\dfrac{5V}{4r}} = \frac{4r}{5}$

Therefore the equivalent resistance between the two opposite corners is $\dfrac{4}{5}\,r$.

Option A is correct.

after removing capacitor branch:(as in steady state capacitor is an open circuit)

• from middle node → one 8Ω to bottom
• from middle → 4Ω → right node
• right node → two 8Ω in parallel to bottom

step 1: right side

$8||8=4\Omega$

step 2: path via right:

$4+4=8\Omega$

so from middle node to bottom we have:

$8\Omega\text{ (direct)}\parallel8\Omega\text{ (via right)}$

R=4Ω

step 3: total resistance:

$R_{total}=1+4=5Ω$

step 4: current:

$I=\frac{10}{5}=2A$

current splits equally in two branches so answer is 1A

From the PV diagram, the two curved paths are adiabatic (their shape indicates $PV^{\gamma}=\text{constant}$, and the process from $P_1$ to $P_2$​ at  $V=1m^3$ is vertical, i.e. an isochoric process.

Since volume is constant, work done is zero:

W=0

So heat supplied equals change in internal energy:

$Q=nC_v(T_2-T_1)$

Using ideal gas law,

$T_1=\frac{P_1V}{nR}=\frac{21.7\times1}{10\times8.3}$

$T_2=\frac{P_2V}{nR}=\frac{30\times1}{10\times8.3}$

Thus,

$T_2−T_1=\frac{\left(30−21.7\right)}{83}=\frac{8.3}{83}=0.1K$

Therefore,

$Q=10\times21\times0.1$

$Q=21J$

So,

$α=21$

let $C_0=\frac{\varepsilon_0A}{d}$

case (A)

top layer (d/2, full area, $k_1$​):

$C_t=\frac{k_1\varepsilon_0A}{d/2}=\frac{2k_1\varepsilon_0A}{d}$

bottom layer (d/2 split):

$C_{b1}=\frac{k_1\varepsilon_0(A/2)}{d/2}=\frac{k_1\varepsilon_0A}{d}$

$C_{b2}=\frac{k_2\varepsilon_0(A/2)}{d/2}=\frac{k_2\varepsilon_0A}{d}$

parallel:

$C_b=\frac{(k_1+k_2)\varepsilon_0A}{d}$

series:

$C_A=\frac{\varepsilon_0A}{d}\cdot\frac{2k_1(k_1+k_2)}{3k_1+k_2}$

case (B)

top: $k_2$

$C_t=\frac{2k_2\varepsilon_0A}{d}$

bottom same as before:

$C_b=\frac{(k_1+k_2)\varepsilon_0A}{d}$

$C_B=\frac{\varepsilon_0A}{d}\cdot\frac{2k_2(k_1+k_2)}{k_1+3k_2}$

case (C)

top split:

$C_t=\frac{(k_1+k_2)\varepsilon_0A}{d}$

bottom split:

$C_b=\frac{(k_1+k_2)\varepsilon_0A}{d}$

series of equal:

$C_C=\frac{1}{2}\cdot\frac{(k_1+k_2)\varepsilon_0A}{d}$

now compare (given $k_1>k_2$):

$C_A>C_C>C_B$

Moment of Inertia about the Pivot ($I_P$):

The rod is pivoted at a distance of $L/3$ from the bottom. The center of mass (CM) of the rod is at its midpoint, $L/2$ from the bottom.

The distance ($d$) between the pivot and the center of mass is:

$d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$

Using the Parallel Axis Theorem:

$I_P = I_{cm} + Md^2$

$I_P = \frac{1}{12}ML^2 + M\left(\frac{L}{6}\right)^2$

$I_P = \frac{1}{12}ML^2 + \frac{1}{36}ML^2 = \frac{1}{9}ML^2$

Change in Potential Energy ($\Delta PE$):

When the rod falls from the vertical position to the horizontal position, the center of mass descends by a vertical height equal to its initial distance from the pivot.

$\Delta h = d = \frac{L}{6}$

$\Delta PE = Mg\Delta h = Mg\left(\frac{L}{6}\right)$

Conservation of Energy:

$\Delta PE = \Delta KE_{rot}$

$Mg\left(\frac{L}{6}\right) = \frac{1}{2} I_P \omega^2$

$Mg\frac{L}{6} = \frac{1}{2} \left(\frac{1}{9}ML^2\right) \omega^2$

$\omega = \sqrt{\frac{3g}{L}}$

Magnetic field on the axis of a circular current-carrying loop, $B = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$

Taking direction from P to Q as positive $\hat{i}$: $x = r$

For loop P:

$I_P = I$

$\vec{B}_P = -\frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} \hat{i}$

$\vec{B}_P = -\frac{\mu_0 I r^2}{2(2r^2)^{3/2}} \hat{i}$

$\vec{B}_P = -\frac{\mu_0 I}{2 \cdot 2\sqrt{2} r} \hat{i}$

$\vec{B}_P = -\frac{\mu_0 I}{4\sqrt{2}r} \hat{i}$

For loop Q:

$I_Q = 4I$

$\vec{B}_Q = \frac{\mu_0 (4I) r^2}{2(r^2 + r^2)^{3/2}} \hat{i}$

$\vec{B}_Q = \frac{4\mu_0 I}{4\sqrt{2}r} \hat{i}$

Net magnetic field at O: $\vec{B}_{\text{net}} = \vec{B}_P + \vec{B}_Q$

$\vec{B}_{\text{net}} = \left(-\frac{\mu_0 I}{4\sqrt{2}r} + \frac{4\mu_0 I}{4\sqrt{2}r}\right) \hat{i}$

$\vec{B}_{\text{net}} = \frac{3\mu_0 I}{4\sqrt{2}r} \hat{i}$

For a point source,

Intensity varies as

$I\propto\frac{1}{r^2}$

If volume of spherical detector increases 8 times,

$V'=8V$

For sphere,

$V=\frac{4}{3}\pi r^3$

so

$r'^3=8r^3$

$r'=2r$

Now intensity becomes

$I'=\frac{1}{(2r)^2}$

$=\frac{1}{4}\cdot\frac{1}{r^2}$

So

$I'=\frac{I}{4}$

Hence intensity becomes one-fourth of original.

Take the chain and split it into two equal halves. Because both supports are at the same level, the situation is symmetric.

Now focus on only one half of the chain.

For this half:

• its weight acts downward and equals $\frac{mg}{2}$​
• the tension at the support is $T_s$ making an angle $30^{\circ}$ with the horizontal
• at the lowest point, the tension is horizontal (no vertical component there)

Now apply equilibrium.

Vertical direction:

Only the support tension provides an upward component, so it must balance the weight of half the chain:

$T_s\sin30^{\circ}=\frac{mg}{2}$

$Since\ \sin30^{\circ}=\frac{1}{2}$

$T_s\cdot\frac{1}{2}=\frac{mg}{2}$

$\Rightarrow T_s=mg$

Now look at horizontal direction.

The horizontal component of tension is the same everywhere in the chain. That means the tension at the lowest point is equal to the horizontal component of TsT_sTs​:

$T_{lowest}=T\cos30^{\circ\ }$

Substitute $T_s=mg\ and\cos30^{\circ}=\frac{\sqrt{3}}{2}$

$T_{\text{lowest}}=mg\cdot\frac{\sqrt{3}}{2}$

So the tension at the lowest point is

$T_{\text{lowest}}=mg\cdot\frac{\sqrt{3}}{2}$

We need to find the least count of the vernier callipers.

50 vernier scale divisions (VSD) = 48 main scale divisions (MSD)

1 MSD = 0.05 mm

Find the value of 1 VSD:

$1 \text{ VSD} = \frac{48}{50} \text{ MSD} = \frac{48}{50} \times 0.05 = 0.048 \text{ mm}$

Calculate the least count:

$\text{Least Count} = 1 \text{ MSD} - 1 \text{ VSD} = 0.05 - 0.048 = 0.002 \text{ mm}$

Therefore, the least count is Option 3: 0.002 mm.

We need to find x where the ratio of lengths of closed pipe to open pipe is 5/x.

Key formulas:

For a closed organ pipe, the nth harmonic frequency (only odd harmonics): $f_n = \frac{nv}{4L_c}$ where n = 1, 3, 5, ...

For an open organ pipe, the nth harmonic: $f_n = \frac{nv}{2L_o}$ where n = 1, 2, 3, ...

Fifth harmonic of closed pipe = First harmonic of open pipe.

Fifth harmonic of closed pipe: $f = \frac{5v}{4L_c}$

First harmonic of open pipe: $f = \frac{v}{2L_o}$

Setting them equal:

$\frac{5v}{4L_c} = \frac{v}{2L_o}$

$\frac{5}{4L_c} = \frac{1}{2L_o}$

$\frac{L_c}{L_o} = \frac{5 \times 2}{4} = \frac{10}{4} = \frac{5}{2}$

So $\frac{L_c}{L_o} = \frac{5}{2}$, which means x = 2.

Therefore, the value of x is Option 3: 2.

We need to find the focal length of a concave mirror that produces a 3x magnified real image with object-image distance of 40 cm.

Magnification: m = -3 (real image is inverted, so negative for a mirror)

Distance between object and image = 40 cm

Set up equations:

For a mirror: $m = -\frac{v}{u}$

$-3 = -\frac{v}{u} \implies v = 3u$

Using sign convention (both u and v are negative for concave mirror with real image):

Let u = -x (object distance), then v = -3x (image distance, real image).

Distance between object and image: |v - u| = |-3x - (-x)| = |-2x| = 2x = 40 cm

So x = 20 cm, u = -20 cm, v = -60 cm.

Find focal length using mirror formula:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-60} + \frac{1}{-20} = -\frac{1}{60} - \frac{1}{20} = -\frac{1+3}{60} = -\frac{4}{60} = -\frac{1}{15}$

$f = -15$ cm

Therefore, the focal length is Option 2: -15 cm.

The galvanometer behaves like a resistor $R_g = 100\Omega$ that can carry a maximum (full-scale) current of $I_g = 1\text{ mA} = 0.001\text{ A}$. To measure larger currents we connect a small resistance $R_s$ in parallel with the galvanometer. This combination is called a shunt-type ammeter.

Let the required full-scale current of the ammeter be $I = 5\text{ mA} = 0.005\text{ A}$. When this current enters the parallel combination, part $I_g$ flows through the galvanometer and the remainder $I_s$ flows through the shunt:

$I = I_g + I_s$

Because the galvanometer and the shunt are in parallel, the voltage across each branch is the same:

$I_g\,R_g = I_s\,R_s$ $-(1)$

From the current relation, the shunt current is

$I_s = I - I_g = 0.005 - 0.001 = 0.004\text{ A}$

Substitute $I_g$, $I_s$ and $R_g$ into equation $-(1)$ to obtain $R_s$:

$R_s = \frac{I_g\,R_g}{I_s} = \frac{0.001 \times 100}{0.004} = \frac{0.1}{0.004} = 25\Omega$

Therefore, the resistance that must be connected in parallel with the galvanometer to convert it into an ammeter of full-scale 5 mA is $25\Omega$.

Option C is correct.

Convex lens with power +5D (focal length f = 1/5 m = 20 cm). Check readings of (u, v) pairs.

Apply lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for each pair:

With $f = 20$ cm, $u$ is negative (real object), $v$ is positive (real image).

For $P_1$: $(u, v) = (-25, 96)$. $\frac{1}{96} + \frac{1}{25} = 0.0104 + 0.04 = 0.0504 \approx 0.05 = 1/20$. Correct.

For $P_2$: $(u, v) = (-30, 62)$. $\frac{1}{62} + \frac{1}{30} = 0.0161 + 0.0333 = 0.0495 \approx 0.05$. Approximately correct.

For $P_3$: $(u, v) = (-35, 37)$. $\frac{1}{37} + \frac{1}{35} = 0.0270 + 0.0286 = 0.0556 \neq 0.05$. INCORRECT. (Expected v: $\frac{1}{v} = 1/20 - 1/35 = 3/140$, $v = 46.67$).

For $P_4$: $(u, v) = (-45, 35)$. $\frac{1}{35} + \frac{1}{45} = 0.0286 + 0.0222 = 0.0508 \approx 0.05$. Approximately correct.

For $P_5$: $(u, v) = (-50, 32)$. $\frac{1}{32} + \frac{1}{50} = 0.03125 + 0.02 = 0.05125 \approx 0.05$. Approximately correct.

Only $P_3$'s reading is significantly off.

The correct answer is Option D: Readings recorded by $P_3$ person are incorrect.

For the ideal gas, $C_v = C_p - R = \dfrac{7}{2}R - R = \dfrac{5}{2}R$. At constant volume, $Q = nC_v\Delta T$. With $Q = 300$ J, $\Delta T = 50 - 20 = 30$ K, and $R = 8.314$ J/mol K:

$300 = n \times \dfrac{5}{2} \times 8.314 \times 30 \implies n = \dfrac{300}{623.55} \approx 0.481$ mol.

The answer is $\boxed{481}$.

We need to find the ratio $I_1/I_2$ of moments of inertia of a solid cylinder and a smaller co-centric cylinder carved from it.

Moment of inertia of a solid cylinder about its axis:

$I = \frac{1}{2}MR^2$

For the large cylinder:

Mass: $M_1 = \rho \pi R^2 L$

$I_1 = \frac{1}{2}M_1 R^2 = \frac{1}{2}\rho \pi R^2 L \times R^2 = \frac{1}{2}\rho \pi R^4 L$

For the small cylinder:

Radius = R/3, Length = L/2

Mass: $M_2 = \rho \pi (R/3)^2 (L/2) = \rho \pi R^2 L / 18$

$I_2 = \frac{1}{2}M_2 (R/3)^2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times \frac{R^2}{9} = \frac{\rho \pi R^4 L}{324}$

Ratio:

$\frac{I_1}{I_2} = \frac{\frac{1}{2}\rho \pi R^4 L}{\frac{\rho \pi R^4 L}{324}} = \frac{324}{2} = 162$

Therefore, $I_1/I_2 =$ 162.

We begin by noting that we need to find x where the work done in blowing a soap bubble from diameter 7 cm to 14 cm is (15000 - x) μJ.

We recall that a soap bubble has two surfaces, so the formula for the work done is:

$W = T \times \Delta A \times 2 = 2T \times 4\pi(R_2^2 - R_1^2)$

$W = 8\pi T(R_2^2 - R_1^2)$

Here, T = 0.04 N/m.

Also, $R_1 = 3.5$ cm = 0.035 m, $R_2 = 7$ cm = 0.07 m.

We take $\pi = 22/7$.

Substituting into the formula gives:

$W = 8 \times \frac{22}{7} \times 0.04 \times (0.07^2 - 0.035^2)$

$= 8 \times \frac{22}{7} \times 0.04 \times (0.0049 - 0.001225)$

$= 8 \times \frac{22}{7} \times 0.04 \times 0.003675$

$= 8 \times \frac{22}{7} \times 0.000147$

$= 8 \times 22 \times \frac{0.000147}{7}$

$= 8 \times 22 \times 0.000021$

$= 8 \times 0.000462$

$= 0.003696 \text{ J} = 3696 \text{ μJ}$

Thus, $15000 - x = 3696$

and $x = 15000 - 3696 = 11304$

Therefore, x = 11304.