Physics 2005 Jamb Examination
Question 31
Magnetic force on a moving charge is \(\mathbf{F}=q\mathbf{v}\times\mathbf{B}\). For two particles with velocities in the same direction and a uniform magnetic field perpendicular to that velocity, the force direction depends on the sign of \(q\).
If the particles are deflected in opposite directions the only consistent explanation is that their charges have opposite signs (one positive, one negative). Hence option A is correct.
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Question 32
The resistance of a piece of wire of length 20m has crossed-sectional area 8 x 10−6m2 is?
[Resistivity of the wire = 4 x 10−7 Ω m]
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Question 33
The figure above shows a plot of angles of deviation through a glass prism when light is incident at \(\theta\) degrees on the prism. The incident angle that produces the minimum deviation is
From the given graph, when the angle of minimum deviation = 18º, the angle of incidence = 25º
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Question 34
The value of F in the figure above when in equilibrium is
= 20N
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Question 35
In the figure above, the coefficient of static friction is 0.5. If the tension in the string is T, the acceleration of the system in motion is
Normal reaction R = 3 kg x 10 = 30N
coefficient of friction , μ = 0.5
∴ acting frictional force = μR = 30 x 0.5
Pulling/sliding force F = 6kg x 10 = 60N.
∴ Net downward force = ma = 60 - 15.0
=> (3 + 6)a = 45
∴ 9a = 45
a = 5m/s2
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Question 36
If the system above is in equilibrium, the tension in the string Q is
If the string is in equilibrium, the the tension in the string Q = 4N + 5N = 9N
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Question 37
In the figure above, the work done by the force of 100 N inclined at an angle of 60° to the object dragged horizontally to a distance of 8 m is
Work done by a force at angle \(\theta\) to the displacement is \(W = F\cos\theta \times d\).
Substitute \(F=100\,\text{N}\), \(\theta=60°\), \(d=8\,\text{m}\):
\[W = 100\cos60^\circ \times 8 = 100\times\tfrac{1}{2}\times 8 = 400\,\text{J}.\]
Therefore option D (400 J) is correct.
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Question 38
In the circuit diagram of a transistor above, the terminal P, Q and R represent
The transistor terminal labels (P, Q, R) are the EMITTER, BASE and COLLECTOR terminals respectively.
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Question 39
Balance mass numbers: initial mass \(=235+1=236\). Products shown: \(144 + 90 = 234\). Remaining mass = \(236-234 = 2\). Those two identical particles therefore each have mass number 1, so they are neutrons (\(^1_0\text{n}\)).
This is the classic fission pattern: uranium-235 absorbs a neutron and often splits into two fission fragments plus additional neutrons. Therefore X is a neutron (option D).
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Question 40
Two liquids L1 and L2 are contained in a U-tube. The height and the density of L1 are 8 cm and 1030 kg m-3 respectively. If the density of L2 is 800 kg m-3, its height measured from the same level is
Hydrostatic equilibrium at the common datum level gives \(\rho_1 h_1 = \rho_2 h_2\).
Assuming \(\rho_1 = 1030\,\text{kg m}^{-3}\) (likely a typographical fix from 103), \(h_1 = 8\,\text{cm}\), and \(\rho_2 = 800\,\text{kg m}^{-3}\),
\[h_2 = \dfrac{\rho_1 h_1}{\rho_2} = \dfrac{1030\times 8}{800} = \dfrac{8240}{800} = 10.3\,\text{cm} \approx 10\,\text{cm}.\]
Therefore option B (10 cm) is the best match. If you want me to use the literal density 103 kg m^{-3} I will recalc and update the row.
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