Question 1
For the flow of a fluid through a pipe, if the Reynolds number ($R_e$) lies strictly between 1000 and 2000, the nature of the flow is classified as:
✅ Correct Answer: Unsteady and transitioning
Explanation: According to NCERT Fluid Mechanics, if Reynolds number $R_e < 1000$, the flow is steady or laminar. If $R_e > 2000$, the flow becomes completely turbulent. Agar $R_e$ value 1000 aur 2000 ke beech mein ho, toh flow unsteady ho jata hai aur laminar se turbulent mein transition karta hai.
Question 2
If the ultimate tensile strength point and the fracture point are located extremely close to each other on a material's stress-strain curve, the material is classified as:
✅ Correct Answer: Brittle
Explanation: NCERT Mechanical Properties of Solids states that if the ultimate strength point and the fracture point are close together on the graph, the material breaks almost immediately after crossing its elastic limit. Aise materials ko hum brittle kehte hain (like glass). Agar ye points door ho, toh material ductile hota hai.
Question 3
The core of an electromagnet is typically made using a ferromagnetic material (like soft iron) that possesses:
✅ Correct Answer: High permeability and low retentivity
Explanation: Electromagnet banane ke liye hume aisa material chahiye jo switch on karte hi easily magnetize ho jaye (High Permeability) aur switch off karte hi apna magnetism turant lose kar de (Low Retentivity). Soft iron completely fits this requirement, preventing permanent residual magnetism.
Question 4
When unpolarized light is incident from air onto a plane glass surface at Brewster's angle ($i_p$), the angle between the completely polarized reflected ray and the partially polarized refracted ray is:
✅ Correct Answer: $90^\circ$
Explanation: According to Brewster's Law, when light is incident at the polarizing angle $i_p$, the reflected ray is completely plane-polarized. At this exact condition, the reflected ray and the refracted ray are perpendicular to each other, making the angle between them exactly $90^\circ$.
Question 5
To get a digital output of 1 from a two-input NAND gate, the inputs $A$ and $B$ must be:
✅ Correct Answer: Both B and C are correct
Explanation: The boolean expression for a NAND gate is $Y = \overline{A \cdot B}$.
If $A=1, B=1 \implies Y = \overline{1} = 0$
If $A=1, B=0 \implies Y = \overline{0} = 1$
If $A=0, B=1 \implies Y = \overline{0} = 1$
Isliye inputs dynamic ho sakte hain, and both conditions yield an output of 1.
Question 6
A point charge $+q$ is placed exactly at the open rim/mouth of a hemispherical cup of radius $R$. The total electric flux passing through the curved surface of this cup is:
✅ Correct Answer: $\frac{q}{2\varepsilon_0}$
Explanation: Gauss's law apply karne ke liye hume symmetric closed surface chahiye. Hum ek identical imaginary hemisphere upar rakh kar charge ko complete enclose karenge. Ab total closed sphere se flux $\frac{q}{\varepsilon_0}$ hoga. By symmetry, niche wale original hemisphere se safe flux half ho jayega, target flux = $\frac{q}{2\varepsilon_0}$.
Question 7
Taking end correction into account, the fundamental frequency of a closed organ pipe of length $L$ and internal radius $r$ is given by:
✅ Correct Answer: $\nu=\frac{v}{4(L+0.6r)}$
Explanation: In a closed organ pipe, reflection happens slightly outside the open end. Effective length is taken as $L' = L+e$, where end correction for one open end is $e=0.6r$. Fundamental frequency ka standard format $\nu = \frac{v}{4L'}$ hota hai, which yields $\nu = \frac{v}{4(L+0.6r)}$.
Question 8
According to Bohr's atomic model, the ratio of the magnetic dipole moment ($M$) of an orbital electron to its orbital angular momentum ($L$) is a universal constant known as the gyromagnetic ratio. Its value is:
✅ Correct Answer: $\frac{e}{2m}$
Explanation: The magnetic moment of a revolving electron is given by $M = \frac{evr}{2}$ and its angular momentum is $L=mvr$. Dividing the two equations gives: $\frac{M}{L} = \frac{evr/2}{mvr} = \frac{e}{2m}$. This value is independent of the orbit number and represents the gyromagnetic ratio.
Question 9
Assertion (A): The coefficient of kinetic friction ($\mu_k$) is always slightly less than the coefficient of static friction ($\mu_s$) for a given pair of surfaces.
Reason (R): Once the actual motion starts, the microscopic contact points between the surfaces do not get enough time to interlock deeply.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: Static friction acts when the body is at rest and regular interlocking is deep. Jab motion start ho jata hai, tab contact points ko lock hone ka perfect time nahi milta, behaves dynamically. Hence kinetic friction becomes slightly smaller than maximum static friction ($\mu_k < \mu_s$). Reason explicitly satisfies the assertion.
Question 10
Assertion (A): A diamond sparkles with exceptional brilliance when cut with specific proper angles.
Reason (R): The critical angle for a diamond-air interface is extremely large, which traps most of the entering light inside through multiple total internal reflections.
✅ Correct Answer: (A) is true but (R) is false.
Explanation: Assertion operates perfectly; diamond cutting enhances brilliance. However, the refractive index of diamond is very high ($\mu \approx 2.42$). Since $\sin \theta_c = 1/\mu$, a high $\mu$ values means the critical angle $\theta_c$ is actually extremely small ($\theta_c \approx 24.4^\circ$), not large! Light easily enters but finds it difficult to escape. Therefore, Reason is false.
Question 11
Consider the following two statements regarding nuclear physics:
Statement I: The strong nuclear force is highly charge-independent, meaning it acts equally between a proton-proton, neutron-neutron, and proton-neutron.
Statement II: For a stable nucleus, the total mass of the nucleus is always strictly greater than the sum of the individual masses of its constituent nucleons when separated.
✅ Correct Answer: Statement I is correct but Statement II is incorrect.
Explanation: Statement I correct hai: Nuclear force charge par depend nahi karta, it's purely a strong interaction. Statement II incorrect hai: Stable nucleus ka mass humesha constituent nucleons ke isolated masses ke sum se kam hota hai. Yeh missing mass energy form ($E = \Delta m \cdot c^2$) mein release hota hai jise binding energy kehte hain.
Question 12
Consider the following two statements regarding thermal physics:
Statement I: According to Kirchhoff's law of thermal radiation, good absorbers of a particular wavelength are also good emitters of that same wavelength.
Statement II: A perfectly blackbody emits radiation only when its internal temperature matches the surrounding ambient temperature.
✅ Correct Answer: Statement I is correct but Statement II is incorrect.
Explanation: Statement I is correct: This is the exact fundamental formulation of Kirchhoff's radiation law ($a_\lambda = e_\lambda$). Statement II is incorrect: According to Stefan's Law, any body above $0\text{ K}$ emits thermal radiation continuously, completely independent of the surrounding temperature.
Question 13
A vehicle of mass $m$ is moving on a smoothly banked circular track of radius $R$ with a banking angle $\theta$. If the coefficient of static friction between the tires and the road surface is $\mu_s$, the maximum safe speed ($v_{\max}$) to avoid slipping upwards is:
✅ Correct Answer: $v_{\max} = \sqrt{gR \left(\frac{\mu_s + \tan\theta}{1-\mu_s \tan\theta}\right)}$
Explanation: When the vehicle moves at maximum safe velocity, friction acts downwards along the banked incline track. Resolving forces vertically and horizontally gives the classic NTA expression: $v_{\max} = \sqrt{gR \left(\frac{\mu_s + \tan\theta}{1-\mu_s \tan\theta}\right)}$. Option C represents the minimum safe speed to prevent sliding downwards.
Question 14
In a standard meter bridge setup, the balancing length from the left end is found to be $40\text{ cm}$ when unknown resistors $X$ and $Y$ are placed in the left and right gaps respectively. When a shunt resistor of $30\,\Omega$ is connected across $Y$, the balancing length shifts to $50\text{ cm}$. Find the value of resistance $X$.
✅ Correct Answer: $10\,\Omega$
Explanation: Case 1: $\frac{X}{Y} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3} \implies Y = 1.5X$.
Case 2: Shunt across $Y$ means new parallel resistance $Y' = \frac{30Y}{30+Y}$.
$\frac{X}{Y'} = \frac{50}{50} = 1 \implies X = Y' = \frac{30Y}{30+Y}$.
Substitute $Y = 1.5X$: $X = \frac{30(1.5X)}{30+1.5X} \implies 30 + 1.5X = 45 \implies 1.5X = 15 \implies X = 10\,\Omega$.
Question 15
From a uniform circular disc of radius $R$, a small circular hole of radius $R/2$ is cleanly scooped out. The center of this hole lies at a distance of $R/2$ from the center of the original complete disc. The displacement of the center of mass of the remaining structural body from the original center is:
✅ Correct Answer: $-\frac{R}{6}$
Explanation: Let the surface mass density be $\sigma$. Mass of original disc $M = \sigma \cdot \pi R^2$ centered at $(0,0)$. Mass of removed portion $m = \sigma \cdot \pi (R/2)^2 = \frac{M}{4}$ centered at $(R/2,0)$.
Using the negative mass concept for cutouts: $x_{\text{com}} = \frac{M \cdot 0 - m \cdot x_1}{M-m} = \frac{-(M/4)(R/2)}{M - M/4} = \frac{-MR/8}{3M/4} = -\frac{R}{6}$.
The negative sign shows it shifts in the opposite direction of the cutout center.
Question 16
The velocity-time ($v-t$) graph of an object moving along a linear track is a straight line sloping uniformly downwards with a negative slope, crossing the time axis. The corresponding acceleration-time ($a-t$) graph is best represented by:
✅ Correct Answer: A straight line parallel to the time axis in the negative region
Explanation: The slope of a velocity-time graph represents the instantaneous acceleration ($a = \frac{dv}{dt}$). Since the $v-t$ graph is a straight line sloping downwards, its slope is a non-zero, constant negative value. Iska matlab acceleration constant hai aur negative region mein hai. So the $a-t$ graph is a horizontal line below the time axis.
Question 17
The escape velocity from the surface of Earth is $v_e$. Consider an imaginary planet whose total mass is 8 times that of Earth and whose radius is twice that of Earth. The escape velocity from the surface of this planet will be:
✅ Correct Answer: $2v_e$
Explanation: The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$. For the new planet: $v_{\text{planet}} = \sqrt{\frac{2G(8M)}{2R}} = \sqrt{4 \times \frac{2GM}{R}} = 2 \cdot \sqrt{\frac{2GM}{R}} = 2v_e$.
Question 18
Two concentric, coplanar circular loops of radii $r_1$ and $r_2$ ($r_1 \gg r_2$) are placed such that their centers perfectly coincide. The mutual inductance ($M$) between these loops scales with their radii as:
✅ Correct Answer: $M \propto \frac{r_2^2}{r_1}$
Explanation: Let a current $I_1$ flow through the outer loop of radius $r_1$. The magnetic field produced at the center is $B_1 = \frac{\mu_0 I_1}{2r_1}$. Since $r_1 \gg r_2$, this field can be assumed uniform over the inner loop. The magnetic flux through the inner loop is: $\phi_2 = B_1 \cdot A_2 = \left(\frac{\mu_0 I_1}{2r_1}\right) \cdot (\pi r_2^2)$. We know $\phi_2 = M \cdot I_1$. Comparing terms gives $M = \frac{\mu_0 \pi r_2^2}{2r_1}$, hence $M \propto \frac{r_2^2}{r_1}$.
Question 19
In a standard Young's Double Slit Experiment (YDSE), the monochromatic light source is completely replaced by a white light source. What configuration is observed on the screen?
✅ Correct Answer: A white central fringe surrounded by a few colored bands, tracking into uniform white illumination.
Explanation: At the exact center of the screen, the path difference is zero ($\Delta x=0$) for all wavelengths present in white light. Therefore, all colors interfere constructively at the center, creating a crisp white central fringe. Move karne par, dynamic wavelengths separate out because fringe width $\beta = \frac{\lambda D}{d}$ is different for different colors, creating a few colored bands before overlapping into uniform white illumination.
Question 20
A block of mass $5\text{ kg}$ is pressed firmly against a rough vertical wall by applying a constant horizontal force of $100\text{ N}$. If the coefficient of static friction between the block and the wall surface is $0.6$, the magnitude of the frictional force acting on the block is: ($g = 10\text{ m/s}^2$).
✅ Correct Answer: $50\text{ N}$
Explanation: Normal force tracking: Horizontal equilibrium means Normal reaction $N = F_{\text{applied}} = 100\text{ N}$. Maximum available static friction capability is $f_{s,\max} = \mu_s N = 0.6 \times 100 = 60\text{ N}$. The downward force pulling the block down is its weight: $W = mg = 5 \times 10 = 50\text{ N}$. Since the pulling force ($50\text{ N}$) is less than the maximum friction threshold ($60\text{ N}$), the block will not move. Static friction matches the applied pulling force to maintain equilibrium, so $f_s = 50\text{ N}$.
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