WB SLST Test 1

Physics Quiz

Physics Quiz

Time: 1 hr

30 questions

1. For any vectors $\vec{A}$ and $\vec{B}$ the scalar $(\vec{A}\times\vec{B}) \cdot \vec{B}$ equals
  • A) $0$
  • B) $\vec{A}\cdot\vec{B}$
  • C) $|\vec{A}\times\vec{B}|$ $|\vec{B}|^{2}$
  • D) None of the above

The expression $(\vec{A}\times\vec{B}) \cdot \vec{B}$ represents a **scalar triple product**. The vector $(\vec{A}\times\vec{B})$ is a new vector that is perpendicular to both $\vec{A}$ and $\vec{B}$.

The dot product of two vectors is zero if they are perpendicular. Since $(\vec{A}\times\vec{B})$ is perpendicular to $\vec{B}$ by definition of the cross product, their dot product is zero.

Therefore, $(\vec{A}\times\vec{B}) \cdot \vec{B} = 0$.

2. For a projectile launched on a level plane with fixed speed, the angle for maximum horizontal range is
  • A) $30^{\circ}$
  • B) $45^{\circ}$
  • C) $60^{\circ}$
  • D) $90^{\circ}$

The horizontal range ($R$) of a projectile launched with initial speed $v_0$ and launch angle $\theta$ is given by the formula:

$$R = \frac{v_0^2 \sin(2\theta)}{g}$$

To maximize the range, the term $\sin(2\theta)$ must be at its maximum value, which is 1. This occurs when the angle $2\theta$ is $90^{\circ}$.

$$2\theta = 90^{\circ} \implies \theta = 45^{\circ}$$

3. A rigid body rotates about a fixed axis with angular acceleration $\alpha$. The tangential acceleration $a_t$ of a point at radius r is
  • A) $a_t = \alpha r$
  • B) $a_t = \frac{\alpha}{r}$
  • C) $a_t = \frac{r}{\alpha}$
  • D) $a_t = \alpha + r$

For a point on a rotating rigid body, the **tangential acceleration** ($a_t$) is the component of acceleration that is tangent to the circular path of the point. It is directly proportional to both the angular acceleration ($\alpha$) of the body and the radius ($r$) of the point's path.

The relationship is given by the formula: $$a_t = \alpha r$$

This is the rotational equivalent of the linear equation $a = \frac{dv}{dt}$.

4. For pure rolling (no slipping) of a wheel of radius R with angular speed $\omega$, the linear speed v of its center is
  • A) $v = \omega R$
  • B) $v = \frac{\omega}{R}$
  • C) $v = \frac{R}{\omega}$
  • D) $v = \omega^{2}R$

In the case of **pure rolling**, the point of contact between the wheel and the surface has zero velocity. This condition ensures that the linear speed ($v$) of the center of mass is directly related to the angular speed ($\omega$) and the radius ($R$) of the wheel.

The relationship is given by the formula: $$v = \omega R$$

This equation holds true only when there is no slipping.

5. Reynolds number Re in pipe flow is used to predict whether flow is laminar or turbulent. Flow is typically laminar when Re is
  • A) $< 2000$
  • B) $2000-4000$
  • C) $> 4000$
  • D) irrelevant to laminar/turbulent transition

The **Reynolds number (Re)** is a dimensionless quantity used to predict fluid flow patterns. It represents the ratio of inertial forces to viscous forces within a fluid.

For flow in a pipe, the transition between laminar (smooth, orderly flow) and turbulent (chaotic, disorderly flow) is generally defined by the following ranges:

  • **Laminar flow:** $Re < 2000$
  • **Transition flow:** $2000 < Re < 4000$
  • **Turbulent flow:** $Re > 4000$

Therefore, a Reynolds number less than 2000 indicates a typically laminar flow.

6. The excess pressure inside a soap bubble of radius R with surface tension $\gamma$ (compared to outside) is
  • A) $\frac{2\gamma}{R}$
  • B) $\frac{4\gamma}{R}$
  • C) $\frac{\gamma}{R}$
  • D) $0$

A **soap bubble** has two free surfaces, an inner one and an outer one, unlike a liquid drop which has only one. Both surfaces contribute to the excess pressure due to surface tension.

The excess pressure ($\Delta P$) is given by the formula: $$\Delta P = \frac{4\gamma}{R}$$

The excess pressure for a liquid drop, which has only one surface, is $\frac{2\gamma}{R}$.

7. According to Bernoulli's principle for ideal steady streamline flow, if the fluid speed increases along a streamline (height same), the static pressure
  • A) increases
  • B) decreases
  • C) remains same
  • D) none of the above

**Bernoulli's principle** states that for an ideal fluid, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

The equation is: $$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$$

If the height ($h$) remains the same, the term $\rho gh$ is constant. For the total value to remain constant, if the velocity ($v$) increases, the static pressure ($P$) must decrease.

8. For a diffraction grating, if the wavelength $\lambda$ of incident light increases (other parameters fixed), the diffraction angle of a given order generally
  • A) increases
  • B) decreases
  • C) remains same
  • D) becomes zero

The relationship between the diffraction angle ($\theta$), wavelength ($\lambda$), and grating spacing ($d$) is given by the **diffraction grating equation**:

$$d \sin\theta = m\lambda$$

where $m$ is the order of the diffraction maximum. Rearranging for the diffraction angle, we get:

$$\sin\theta = \frac{m\lambda}{d}$$

If $\lambda$ increases while $m$ and $d$ are fixed, the value of $\sin\theta$ must increase. Since $\sin\theta$ is a monotonically increasing function for angles between $0^{\circ}$ and $90^{\circ}$, the diffraction angle $\theta$ also **increases**.

9. Power (in diopters) of a thin lens of focal length f (in metres) is $P=1/f$. Two thin lenses in contact have combined power equal to
  • A) $P_{1}+P_{2}$
  • B) $\frac{P_{1}P_{2}}{P_{1}+P_{2}}$
  • C) $P_{1}P_{2}$
  • D) $|P_{1}-P_{2}|$

When two or more thin lenses are placed in contact, their powers add up directly. This is because the overall effect on light is a simple sum of the individual effects of each lens.

The total power ($P_{eq}$) of the combination is the sum of the individual powers ($P_1, P_2, ...$).

$$P_{eq} = P_1 + P_2$$

This is a fundamental principle of thin lens combinations.

10. For a concave spherical mirror, if an object is placed at the center of curvature (distance = radius of curvature), the image will be:
  • A) virtual, inverted, same size
  • B) real, inverted, same size, at the center of curvature
  • C) at infinity
  • D) virtual, same size, inverted

When an object is placed at the **center of curvature (C)** of a concave mirror, the image is formed at the same location. This is a special case in spherical mirror optics.

The image is:

  • **Real:** Formed by the actual intersection of reflected rays.
  • **Inverted:** The image is upside down relative to the object.
  • **Same size:** The magnification is -1.
  • **At C:** The image distance is equal to the object distance, which is the radius of curvature ($R$).
11. A straight wire of length L carrying current I lies perpendicular to a uniform magnetic field B. The magnitude of magnetic force on the wire is
  • A) IB
  • B) $ILB \sin\theta$ (for angle between L and B)
  • C) $I^2LB$
  • D) $IB^2L$

The force on a current-carrying wire in a uniform magnetic field is given by the formula:

$$\vec{F} = I(\vec{L} \times \vec{B})$$

The magnitude of this force is the product of the magnitudes of the vectors and the sine of the angle between them ($\theta$).

$$F = ILB \sin\theta$$

If the wire is perpendicular to the magnetic field, $\theta = 90^\circ$ and $\sin(90^\circ) = 1$, so the force is at its maximum magnitude, $F = ILB$.

12. The energy stored in an inductor L carrying current I is
  • A) $\frac{1}{2}LI^{2}$
  • B) LI
  • C) $LI^{2}$
  • D) $\frac{1}{2}L^{2}I$

An **inductor** stores energy in its magnetic field when a current flows through it. The amount of energy ($E$) stored is proportional to the inductance ($L$) and the square of the current ($I$).

The formula for the stored energy is: $$E = \frac{1}{2}LI^2$$

This is analogous to the kinetic energy formula for a mass, $E = \frac{1}{2}mv^2$, where inductance ($L$) is the electrical equivalent of inertia (mass) and current ($I$) is the electrical equivalent of velocity.

13. Two capacitors $C_1$ and $C_2$ are connected in series. The equivalent capacitance $C_{eq}$ is
  • A) $\frac{C_1C_2}{C_1+C_2}$
  • B) $C_1+C_2$
  • C) $C_1C_2$
  • D) $\frac{C_1+C_2}{C_1C_2}$

When capacitors are connected in **series**, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$

To find $C_{eq}$, we can combine the fractions on the right side:

$$\frac{1}{C_{eq}} = \frac{C_2 + C_1}{C_1C_2}$$

Taking the reciprocal of both sides gives the equivalent capacitance:

$$C_{eq} = \frac{C_1C_2}{C_1+C_2}$$

14. In an RC charging circuit the time constant is $\tau=RC$. After time $t=\tau$ the charge on the capacitor reaches approximately
  • A) 50% of final value
  • B) 63% of final value
  • C) 90% of final value
  • D) 37% of final value

The charging of a capacitor in an RC circuit is described by the equation:

$$Q(t) = Q_{max}(1 - e^{-t/\tau})$$

where $Q(t)$ is the charge at time $t$, $Q_{max}$ is the final charge, and $\tau=RC$ is the time constant.

After one time constant, $t=\tau$:

$$Q(\tau) = Q_{max}(1 - e^{-\tau/\tau}) = Q_{max}(1 - e^{-1})$$

Since $e^{-1} \approx 0.368$, the charge is:

$$Q(\tau) \approx Q_{max}(1 - 0.368) = 0.632 Q_{max}$$

This means the charge on the capacitor reaches approximately **63%** of its final value.

15. In Compton scattering the change in wavelength $\Delta\lambda$ of the photon is maximum when the photon is scattered through angle
  • A) $0^{\circ}$
  • B) $90^{\circ}$
  • C) $180^{\circ}$
  • D) $45^{\circ}$

The change in wavelength ($\Delta\lambda$) in **Compton scattering** is given by the Compton shift formula:

$$\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)$$

where $\theta$ is the scattering angle.

To maximize $\Delta\lambda$, the term $(1 - \cos\theta)$ must be at its maximum. The maximum value of this term occurs when $\cos\theta$ is at its minimum value, which is -1. This happens at an angle of $\theta = 180^{\circ}$.

At $\theta = 180^{\circ}$, the photon is scattered directly backward, and the change in wavelength is maximum.

16. Which of the following radiations is the most penetrating in matter (for comparable energies)?
  • A) Alpha particles
  • B) Beta particles
  • C) Gamma rays
  • D) Low-energy visible photons

The penetrating power of radiation depends on its charge and mass. Here's a quick comparison:

  • **Alpha particles ($\alpha$)**: Are helium nuclei (2 protons, 2 neutrons). They are large, massive, and have a positive charge. They interact strongly with matter and are easily stopped (by a sheet of paper or skin).
  • **Beta particles ($\beta$)**: Are electrons or positrons. They are much smaller and less charged than alpha particles. They can penetrate a few millimeters into matter (e.g., aluminum foil).
  • **Gamma rays ($\gamma$)**: Are high-energy photons (electromagnetic radiation). They have no charge and no mass, so they interact very weakly with matter. This makes them highly penetrating, requiring thick, dense materials like lead or concrete to be stopped.

Therefore, **gamma rays** are the most penetrating.

17. The area of the triangle formed by vectors $\vec{a}=(2,0,1)$ and $\vec{b}=(1,1,0)$ (as two sides) is
  • A) $1 \text{ unit}^2$
  • B) $\frac{\sqrt{3}}{2} \text{ unit}^2$
  • C) $\frac{\sqrt{6}}{2} \text{ unit}^2$
  • D) $\frac{1}{2} \text{ unit}^2$

The area of a triangle formed by two vectors is half the magnitude of their cross product. The cross product of $\vec{a}$ and $\vec{b}$ is:

$$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0-1) + \hat{k}(2-0) = -\hat{i} + \hat{j} + 2\hat{k}$$

The magnitude of this vector is:

$$|\vec{a}\times\vec{b}| = \sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$

The area of the triangle is half of this magnitude:

$$\text{Area} = \frac{1}{2}|\vec{a}\times\vec{b}| = \frac{\sqrt{6}}{2} \text{ unit}^2$$

18. The volume of the parallelepiped formed by $\vec{a}=(1,2,3)$, $\vec{b}=(0,1,2)$, $\vec{c}=(1,0,1)$ is (in cubic units)
  • A) 0
  • B) 1
  • C) 2
  • D) 3

The volume of a parallelepiped formed by three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is the absolute value of their **scalar triple product**, which can be calculated using a determinant.

$$\text{Volume} = |\vec{a} \cdot (\vec{b} \times \vec{c})| = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix}$$

Expanding the determinant:

$$\text{Volume} = 1(1 \cdot 1 - 2 \cdot 0) - 2(0 \cdot 1 - 2 \cdot 1) + 3(0 \cdot 0 - 1 \cdot 1)$$

$$\text{Volume} = 1(1) - 2(-2) + 3(-1) = 1 + 4 - 3 = 2$$

The volume is 2. Therefore, option C is the correct answer.

19. A boat moves with speed $5~m/s$ relative to still water. It is aimed straight across a river of width 120 m. The river flows downstream at $3~m/s$. Where will the boat land (downstream displacement from the point opposite the start)?
  • A) 36 m
  • B) 60 m
  • C) 72 m
  • D) 120 m

This is a classic vector problem. The motion of the boat can be analyzed in two independent components: across the river and downstream.

First, calculate the time ($t$) it takes for the boat to cross the river. The time is determined by the boat's velocity perpendicular to the river flow and the river's width.

$$t = \frac{\text{width}}{\text{boat speed}} = \frac{120~m}{5~m/s} = 24~s$$

Next, calculate the downstream displacement. During the 24 seconds it takes to cross the river, the river's current carries the boat downstream. The downstream displacement ($x$) is the product of the river's speed and the time to cross.

$$x = (\text{river speed}) \times t = 3~m/s \times 24~s = 72~m$$

The boat will land 72 m downstream from the point directly opposite its starting point.

20. For simple harmonic motion $x(t)=A\cos(\omega t)$, the maximum magnitude of acceleration is
  • A) $\omega A$
  • B) $\omega^2 A$
  • C) $\frac{A}{\omega}$
  • D) $\frac{\omega}{A}$

The position of an object in simple harmonic motion (SHM) is given by:

$$x(t) = A\cos(\omega t)$$

To find the velocity, we take the first derivative with respect to time:

$$v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t)$$

To find the acceleration, we take the second derivative:

$$a(t) = \frac{dv}{dt} = -A\omega^2\cos(\omega t)$$

The magnitude of acceleration is $|a(t)| = A\omega^2|\cos(\omega t)|$. The maximum value of $|\cos(\omega t)|$ is 1. Therefore, the maximum magnitude of acceleration ($a_{max}$) is:

$$a_{max} = \omega^2 A$$

21. For a simple pendulum of length l (small oscillations), the period T is approximately
  • A) $2\pi\sqrt{\frac{g}{l}}$
  • B) $2\pi\sqrt{\frac{l}{g}}$
  • C) $\pi\sqrt{\frac{l}{g}}$
  • D) $2\sqrt{\frac{l}{g}}$

The period ($T$) of a **simple pendulum** is the time it takes for one complete oscillation. For small-angle oscillations, the motion is approximately simple harmonic motion.

The formula for the period of a simple pendulum of length $l$ in a gravitational field with acceleration $g$ is:

$$T = 2\pi\sqrt{\frac{l}{g}}$$

This formula shows that the period is independent of the mass of the bob and the amplitude of the oscillation (for small angles).

22. Two cylinders (same mass m and radius R) - one solid, one thin hollow - roll without slipping down the same incline from rest. Which reaches the bottom first?
  • A) Hollow cylinder
  • B) Solid cylinder
  • C) Both tie
  • D) Depends on incline angle only

The cylinder that reaches the bottom first will have a greater final linear velocity and thus a greater final kinetic energy. The total kinetic energy is the sum of translational and rotational kinetic energy:

$$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Using the no-slip condition $v = \omega R$, we can write:

$$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}v^2(m + \frac{I}{R^2})$$

The moment of inertia ($I$) for a solid cylinder is $I_{solid} = \frac{1}{2}mR^2$. The moment of inertia for a thin hollow cylinder is $I_{hollow} = mR^2$. Since $I_{hollow} > I_{solid}$, the hollow cylinder requires more energy to rotate.

For a given change in potential energy, the object that puts more energy into rotation will have less energy left for translation, and thus a lower final linear velocity. Since the solid cylinder has a smaller moment of inertia, it has more energy for translation and will reach the bottom first.

23. A force $\vec{F}$ at a point a distance r from an axis. The magnitude of torque about the axis for force making angle $\theta$ with the lever arm is
  • A) $Fr\cos\theta$
  • B) $Fr\sin\theta$
  • C) $Fr$
  • D) $Fr\tan\theta$

The **torque ($\vec{\tau}$)** is a measure of the force that causes an object to rotate. It is a vector quantity defined by the cross product of the position vector ($\vec{r}$) and the force vector ($\vec{F}$).

$$\vec{\tau} = \vec{r} \times \vec{F}$$

The magnitude of the torque is given by the formula:

$$\tau = |\vec{r}||\vec{F}|\sin\theta = Fr\sin\theta$$

where $\theta$ is the angle between the position vector (lever arm) and the force vector. The torque is caused only by the component of the force that is perpendicular to the lever arm, which is $F\sin\theta$.

24. Work done by gravitational force on an object moving once around a closed path (returning to start) is
  • A) zero
  • B) positive
  • C) negative
  • D) depends on path

The **gravitational force** is a **conservative force**. A key property of a conservative force is that the work done by it on an object moving around any closed path is always zero.

This is because the work done by a conservative force only depends on the initial and final positions of the object, not the path taken. Since the object returns to its starting point, its initial and final positions are the same, and the net work done is zero.

25. The molar heat capacity at constant volume $C_V$ for a monoatomic ideal gas equals
  • A) $\frac{3}{2}R$
  • B) $\frac{5}{2}R$
  • C) $2R$
  • D) $R$

A **monoatomic ideal gas** has only three degrees of freedom, which correspond to translational motion in the x, y, and z directions. Each degree of freedom contributes $\frac{1}{2}RT$ to the internal energy per mole.

The internal energy ($U$) of one mole of a monoatomic ideal gas is:

$$U = 3 \times \frac{1}{2}RT = \frac{3}{2}RT$$

The molar heat capacity at constant volume ($C_V$) is the derivative of internal energy with respect to temperature:

$$C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{d}{dT}\left(\frac{3}{2}RT\right) = \frac{3}{2}R$$

26. The reversible work done by n moles of an ideal gas during an isothermal expansion from volume $V_1$ to $V_2$ at temperature T is
  • A) $nRT$
  • B) $nRT \ln \frac{V_2}{V_1}$
  • C) $\frac{1}{2}nR(T_2-T_1)$
  • D) zero

Work done by an expanding gas is given by the integral of pressure with respect to volume: $$W = \int_{V_1}^{V_2} P dV$$

For an ideal gas, $PV = nRT$, so $P = \frac{nRT}{V}$. For an **isothermal process**, the temperature ($T$) is constant.

$$W = \int_{V_1}^{V_2} \frac{nRT}{V} dV = nRT \int_{V_1}^{V_2} \frac{1}{V} dV = nRT [\ln V]_{V_1}^{V_2}$$

$$W = nRT(\ln V_2 - \ln V_1) = nRT \ln \frac{V_2}{V_1}$$

27. A gas suddenly expands into vacuum (free expansion). For an ideal gas, the change in internal energy $\Delta U$ is
  • A) positive
  • B) negative
  • C) zero
  • D) depends on final volume

A **free expansion** is an expansion of a gas into a vacuum. This process is highly irreversible. According to the First Law of Thermodynamics, $\Delta U = Q - W$, where $Q$ is heat added and $W$ is work done by the system.

In a free expansion, the gas does no work because it expands against a vacuum, meaning there is no external pressure ($W = 0$). Additionally, the process happens so quickly that there is no time for heat exchange with the surroundings ($Q = 0$).

Therefore, for a free expansion of an ideal gas, the change in internal energy is zero: $$\Delta U = 0$$

This implies that the temperature of the ideal gas also remains constant.

28. The total power radiated per unit area by a black body is proportional to
  • A) T
  • B) $T^2$
  • C) $T^3$
  • D) $T^4$ (here T = absolute temperature)

This is a direct application of the **Stefan-Boltzmann Law**. The law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time is directly proportional to the fourth power of the black body's absolute temperature ($T$).

The equation is: $$\frac{P}{A} = \epsilon\sigma T^4$$

where $P$ is power, $A$ is area, $\epsilon$ is emissivity (1 for a perfect black body), and $\sigma$ is the Stefan-Boltzmann constant. Thus, the power radiated per unit area is proportional to $T^4$.

29. In Bernoulli's equation the term representing dynamic pressure is
  • A) $\rho gh$
  • B) $p$
  • C) $\frac{1}{2}\rho v^2$
  • D) $\rho v$

Bernoulli's equation for a streamline is typically written as:

$$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$$

The equation has three pressure-like terms:

  • **Static pressure ($P$)**: The actual thermodynamic pressure of the fluid.
  • **Dynamic pressure ($\frac{1}{2}\rho v^2$)**: The pressure associated with the fluid's kinetic energy due to its motion.
  • **Hydrostatic pressure ($\rho gh$)**: The pressure due to the fluid's weight.

Therefore, the term representing dynamic pressure is $\frac{1}{2}\rho v^2$.

30. Magnetic moment $\vec{\mu}$ of a planar current loop of area A carrying current I is equal to
  • A) $I/A$
  • B) $IA$
  • C) $\frac{I}{A}\hat{n}$
  • D) $I^2A$

The **magnetic moment** ($\vec{\mu}$) of a planar current loop is a vector quantity that describes the strength and orientation of the magnetic source. Its magnitude is defined as the product of the current and the area of the loop.

The magnitude is given by: $$\mu = IA$$

The question asks for a scalar value, and option B represents the magnitude of the magnetic moment. While option C is the vector form, B is the intended answer for the scalar magnitude.

Quiz Results

Total questions: 30

Correct answers: 0

Incorrect answers: 0

Score: 0%

Areas to Focus On:

Based on your incorrect answers, you should review the following topics:

    No comments:

    Post a Comment