Solutions for JEE Advanced 2024 Paper 1 Physics Problem 1,2,3,4,5

Q.1: Dimensionless Quantity Construction

Question: Construct a dimensionless quantity using $e$, ${ϵ}_0$, $h$, and $c$. The expression is $e^{\alpha }{ϵ}_0^{\beta }{h}^{\gamma }{c}^{\delta }$. Determine $(\alpha ,\beta ,\gamma ,\delta )$.

Explanation:
To make the quantity dimensionless, the exponents of fundamental dimensions (Mass $M$, Length $L$, Time $T$, Current $I$) must cancel.

1. Dimensions:

   • $[e]=[IT]$

   • $[{ϵ}_0]=[M^{-1}{L}^{-3}{T}^4{I}^2]$

   • $[h]=[M{L}^2{T}^{-1}]$

   • $[c]=[L{T}^{-1}]$

2. Set up equations for exponents:

   • $\alpha +2\beta =0$ (for $I$)

   • $-\beta +\gamma =0$ (for $M$)

   • $-3\beta +2\gamma +\delta =0$ (for $L$)

   • $\alpha +4\beta -\gamma -\delta =0$ (for $T$)

3. Solve:

   • From $\gamma =\beta$, $\delta =\beta$, and $\alpha =-2\beta$.

   • Option A $(2k,-k,-k,-k)$ satisfies all equations.

Answer: $\boxed{{\displaystyle A}}$

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Q.2: Magnetic Field Line Integral

Question: Find the magnitude of $\int \vec{B}\cdot d\vec{l}$ from $(-\sqrt{3}a,a,0)$ to $(a,a,0)$ for an infinite wire on the $z$-axis.

Explanation:

• Magnetic field: $B=\frac{{\mu }_{0}I}{2\pi r}$ (circumferential).

• Path: Horizontal line at $y=a$. Parametrize $x\in [-\sqrt{3}a,a]$.

• Integral:

  $$<span\; style="background-color:\; white;">\int </span>\stackrel{<span\; style="background-color:\; white;">⃗</span>}{\mathrm{<span\; style="background-color:\; white;">B</span>}}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">d</span>}\stackrel{<span\; style="background-color:\; white;">⃗</span>}{\mathrm{<span\; style="background-color:\; white;">l</span>}}<span\; style="background-color:\; white;">=</span>\frac{<span\; style="background-color:\; white;">-</span>{\mathrm{<span\; style="background-color:\; white;">\mu </span>}}_{\mathrm{<span\; style="background-color:\; white;">0</span>}}\mathrm{<span\; style="background-color:\; white;">I</span>}\mathrm{<span\; style="background-color:\; white;">a</span>}}{\mathrm{<span\; style="background-color:\; white;">2</span>}\mathrm{<span\; style="background-color:\; white;">\pi </span>}}{<span\; style="background-color:\; white;">\int </span>}_{<span\; style="background-color:\; white;">-</span>\sqrt{\mathrm{<span\; style="background-color:\; white;">3</span>}}\mathrm{<span\; style="background-color:\; white;">a</span>}}^{\mathrm{<span\; style="background-color:\; white;">a</span>}}\frac{\mathrm{<span\; style="background-color:\; white;">d</span>}\mathrm{<span\; style="background-color:\; white;">x</span>}}{{\mathrm{<span\; style="background-color:\; white;">x</span>}}^{\mathrm{<span\; style="background-color:\; white;">2</span>}}<span\; style="background-color:\; white;">+</span>{\mathrm{<span\; style="background-color:\; white;">a</span>}}^{\mathrm{<span\; style="background-color:\; white;">2</span>}}}$$

• Solve using $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}(x\mathrm{/}a)$:

  $$\text{<span style="background-color: white;">Result</span>}<span\; style="background-color:\; white;">=</span>\frac{\mathrm{<span\; style="background-color:\; white;">7</span>}{\mathrm{<span\; style="background-color:\; white;">\mu </span>}}_{\mathrm{<span\; style="background-color:\; white;">0</span>}}\mathrm{<span\; style="background-color:\; white;">I</span>}}{\mathrm{<span\; style="background-color:\; white;">24</span>}}$$

Answer: $\boxed{{\displaystyle A}}$

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Q.3: Beads on a Hoop Oscillations

Question: Find the square of the angular frequency ${\omega }^2$ for small oscillations of a charged bead on a hoop.

Explanation:

• Force: Electric repulsion $F=\frac{q^2}{16\pi {ϵ}_0{R}^2{\theta }^2}$.

• Restoring force for small $\theta$:

  $${\mathrm{<span\; style="background-color:\; white;">F</span>}}_{\text{<span style="background-color: white;">tangential</span>}}<span\; style="background-color:\; white;">\approx </span>\frac{{\mathrm{<span\; style="background-color:\; white;">q</span>}}^{\mathrm{<span\; style="background-color:\; white;">2</span>}}\mathrm{<span\; style="background-color:\; white;">\theta </span>}}{\mathrm{<span\; style="background-color:\; white;">32</span>}\mathrm{<span\; style="background-color:\; white;">\pi </span>}{\mathrm{<span\; style="background-color:\; white;">ϵ</span>}}_{\mathrm{<span\; style="background-color:\; white;">0</span>}}{\mathrm{<span\; style="background-color:\; white;">R</span>}}^{\mathrm{<span\; style="background-color:\; white;">3</span>}}}$$

• Angular frequency:

  $${\mathrm{<span\; style="background-color:\; white;">\omega </span>}}^{\mathrm{<span\; style="background-color:\; white;">2</span>}}<span\; style="background-color:\; white;">=</span>\frac{{\mathrm{<span\; style="background-color:\; white;">F</span>}}_{\text{<span style="background-color: white;">tangential</span>}}}{\mathrm{<span\; style="background-color:\; white;">m</span>}\mathrm{<span\; style="background-color:\; white;">R</span>}\mathrm{<span\; style="background-color:\; white;">\theta </span>}}<span\; style="background-color:\; white;">=</span>\frac{{\mathrm{<span\; style="background-color:\; white;">q</span>}}^{\mathrm{<span\; style="background-color:\; white;">2</span>}}}{\mathrm{<span\; style="background-color:\; white;">32</span>}\mathrm{<span\; style="background-color:\; white;">\pi </span>}{\mathrm{<span\; style="background-color:\; white;">ϵ</span>}}_{\mathrm{<span\; style="background-color:\; white;">0</span>}}{\mathrm{<span\; style="background-color:\; white;">R</span>}}^{\mathrm{<span\; style="background-color:\; white;">3</span>}}\mathrm{<span\; style="background-color:\; white;">m</span>}}$$

Answer: $\boxed{{\displaystyle B}}$

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Q.4: Block Motion Under Force $F=-20x+10$

Question: Find position and momentum at $t=\pi \mathrm{/}4\text{s}$.

Explanation:

• Equation of motion:

  $$\frac{{\mathrm{<span\; style="background-color:\; white;">d</span>}}^{\mathrm{<span\; style="background-color:\; white;">2</span>}}\mathrm{<span\; style="background-color:\; white;">x</span>}}{\mathrm{<span\; style="background-color:\; white;">d</span>}{\mathrm{<span\; style="background-color:\; white;">t</span>}}^{\mathrm{<span\; style="background-color:\; white;">2</span>}}}<span\; style="background-color:\; white;">+</span>\mathrm{<span\; style="background-color:\; white;">4</span>}\mathrm{<span\; style="background-color:\; white;">x</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">2</span>}\text{<span style="background-color: white;">\hspace{0.25em}\hspace{0.05em}</span>}<span\; style="background-color:\; white;">⟹</span>\text{<span style="background-color: white;">\hspace{0.25em}\hspace{0.05em}</span>}\mathrm{<span\; style="background-color:\; white;">x</span>}<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">t</span>}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">0.5</span>}\mathrm{<span\; style="background-color:\; white;">cos</span>}<span\; style="background-color:\; white;"></span><span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">2</span>}\mathrm{<span\; style="background-color:\; white;">t</span>}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">+</span>\mathrm{<span\; style="background-color:\; white;">0.5</span>}$$

• Initial conditions: $x(0)=1\text{m}$, $v(0)=0$.

• At $t=\pi \mathrm{/}4$:

  • Position: $x=0.5\text{m}$.

  • Momentum: $p=m\cdot v=-5\text{kg m/s}$.

Answer: $\boxed{{\displaystyle C}}$

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Q.5: Bohr Quantization for Central Force $F(r)=-kr$

Question: Identify correct expressions for $v^2$ and $E$.

Explanation:

• Circular orbit:

  • Centripetal force: $kr=\frac{m{v}^2}{r}⟹v^2=\frac{k{r}^2}{m}$.

  • Angular momentum: $L=n\mathrm{\hslash }⟹r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$.

  • Speed: $v^2=\frac{n\mathrm{\hslash }\sqrt{k}}{\sqrt{m^3}}⟹\boxed{{\displaystyle B}}$.

• Energy: $E=\frac{1}{2}m{v}^2+\frac{1}{2}k{r}^2=k{r}^2$.

Answer: $\boxed{{\displaystyle B}}$

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