Physics 2019 Jamb Examination

Physics 2019 Jamb Examination
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Question 11

Electrons were discovered by
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J.J. Thomson discovered the electron in 1897 through his experiments with cathode ray tubes."Image of J.J. Thomson's cathode ray tube experiment
He demonstrated that cathode rays were composed of previously unknown negatively charged particles, which were later named electrons. Dalton proposed the atomic theory, Chadwick discovered the neutron, and Bohr developed the model of electron shells.
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Question 12

Which of the following is/are the limitations to the Rutherford's atomic models?
I. It is applicable when energy is radiated as electrons are revolving
II. It is applicable when energy is radiated in a continuous mode
III. It is applicable to an atom with only one electron in the other shell
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Rutherford's model had two major flaws based on classical physics. First, a revolving electron is accelerating and should continuously radiate energy (II). Second, as it loses energy, it should spiral into the nucleus, making the atom unstable (I). Classical physics also predicted a continuous spectrum, whereas atoms produce discrete line spectra. Statement III is more specifically a limitation of the Bohr model regarding complex atoms.
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Question 13

Which of the following equations is the correct definition of the reactance of an inductor L?
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Inductive reactance (\(X_L\)) is the opposition that an inductor offers to alternating current. Similar to Ohm's law for resistance (\(R = V/I\)), reactance is defined as the ratio of the peak (amplitude) or RMS voltage to the peak or RMS current. The formula is \( X_L = \frac{V_0}{I_0} \).
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Question 14

Ripple in a power supply unit is caused by
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Ripple is the residual periodic variation of the DC voltage within a power supply which has been derived from an AC source. While rectification converts AC to DC, the resulting wave is pulsating. Capacitors are used to smooth this, but under a heavy load, the capacitor discharges faster between peaks, leading to an increase in ripple voltage.
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Question 15

When the temperature of a liquid is increased, its surface tension?
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Surface tension is caused by cohesive forces between liquid molecules. As temperature increases, the kinetic energy of the molecules increases, which weakens these intermolecular forces. Consequently, the surface tension of the liquid decreases.
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Question 16

The distance between an object and its real image in a convex lens is 40cm. If the magnification of the image is 3, calculate the focal length of the lens
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For a real image, the distance between object and image is \( L = u + v = 40 \; \text{cm} \).
Magnification \( M = \frac{v}{u} = 3 \), so \( v = 3u \).
Substituting into the distance equation:\[ u + 3u = 40 \Rightarrow 4u = 40 \Rightarrow u = 10 \; \text{cm} \]Then \( v = 3(10) = 30 \; \text{cm} \).
Using the lens formula:\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]\[ \frac{1}{f} = \frac{1}{10} + \frac{1}{30} = \frac{3 + 1}{30} = \frac{4}{30} \]\[ f = \frac{30}{4} = 7.5 \; \text{cm} \]
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Question 17

A ray of light passes through the centre of curvature of a concave mirror and strikes the mirror. At what angle is the ray reflected?
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Any ray of light passing through the centre of curvature (C) of a spherical mirror hits the surface normally (at 90° to the tangent). According to the law of reflection, the angle of incidence is 0° relative to the normal. Therefore, the angle of reflection is also 0°, causing the ray to reflect back along its own path.
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Question 18

The pin-hole camera produces a less sharply defined image when the
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In a pin-hole camera, a smaller hole allows only a narrow beam of light from each point on the object to reach the screen, creating a sharp image. If the pin-hole is made larger, multiple beams of light from the same point on the object overlap on the screen, resulting in a blurred or less sharply defined image.
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Question 19

A straight wire 15cm long, carrying a current of 6.0A is in a uniform field of 0.40T. What is the force on the wire when it is at right angle to the field
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The magnetic force on a current-carrying wire is given by the formula:\[ F = BIl \sin \theta \]Given:
\( B = 0.40 \; \text{T} \)
\( I = 6.0 \; \text{A} \)
\( l = 15 \; \text{cm} = 0.15 \; \text{m} \)
\( \theta = 90^\circ \) (right angle, so \( \sin 90^\circ = 1 \))
\[ F = 0.40 \times 6.0 \times 0.15 \times 1 \]\[ F = 2.4 \times 0.15 = 0.36 \; \text{N} \]
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Question 20

The pitch of a screw jack is 0.45cm and the arm is 60cm long. If the efficiency of the Jack is 75/Ï€ %, calculate the mechanical advantage.
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First, find the Velocity Ratio (V.R.) of the screw jack:\[ V.R. = \frac{2\pi L}{p} \]where \( L = 60 \; \text{cm} \) and \( p = 0.45 \; \text{cm} \).\[ V.R. = \frac{2\pi \times 60}{0.45} = \frac{120\pi}{0.45} = \frac{12000\pi}{45} = \frac{800\pi}{3} \]Efficiency \( \eta = \frac{75}{\pi}\% = \frac{75}{100\pi} \).
Since \( M.A. = \eta \times V.R. \):\[ M.A. = \frac{75}{100\pi} \times \frac{800\pi}{3} \]\[ M.A. = \frac{75}{3} \times \frac{800}{100} = 25 \times 8 = 200 \]
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Physics 2019 Jamb Examination
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