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The lowest note (fundamental note) emitted by any stretched string has a fundamental frequecy fo = V/2l = 40Hz The 1st overtone = 2fo = 2 x 40 = 80Hz 2nd overtone = 3fo = 3 x 40 = 120Hz 3rd overtone = 4fo = 4 x 40 = 160Hz 4th overtone = 5fo = 5 x 40 = 200Hz Overtones between 40Hz and 180Hz are 80Hz, 120Hz, 160Hz = 3 Overtones

The image of an object placed in front of a plane mirror is always as far behind the mirror as the object is in front of the mirror. Thus when the object is 4m in front of the mirror, the image will be 4m behind the mirror. But when the mirror is moved in towards the man, (that is the man is now 3m from the mirror), the image distance will again decrease by 1m from behind the mirror. Thus the new image position is 3m behind the mirror. Therefore, the distance between the man and his new image is 6m.

Since diameter of the sphere = 20cm, then, the radius of curvature = 10cm Focal length, f = \(\frac{R}{2} = \frac{10}{2}\) = 5cm The inside of a hollow metal sphere forms a concave mirror.

Refractive index = \(\frac{Sin i}{Sin r}\) = \(\frac{\text{Vel. of light in air}}{\text{vel. of light in glass}}\) = \(\frac{Sin i}{Sin 30^o}\)= \(\frac{3.0 \times 10^8}{2.0 \times 10^8}\) = Sin i = \(\frac{(3.0 \times 10^8 \times Sin 30^o)}{2.0 \times 10^8}\) = Sin i = \(\frac{(3.0 \times 10^8 \times 0.5)}{2.0 \times 10^8}\) = \(\frac{1.5}{2}\) = 0.75

An astronomical telescope is in normal adjustment when the final image is at infinity. This means: a) Eye is relaxed (no accommodation) b) Image from objective is at focal point of eyepiece c) Maximum comfortable magnification for distant viewing

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option D.

Since all the cells are connected in parallel, the effective e.m.f. is just that of one single cell. Which means e.m.f. = 1.5V However, the effective internal resistance adds as parallel arrangement given as 1/r = 1/r1 + 1/r2 + 1/r3 + 1/r4 = 1/4 + 1/4 + 1/4 + 1/4 = 1\Omega ∴ r = 1\Omega Answer = 1.5V, 1\Omega

The maximum current drawn by the cooker is given by; I = \(\frac{P}{V}\) ∴ I = \(\frac{\text{3000}}{200}\) I = 15A The fuse rating = I \(\times\) 125% (the fuse is rated at 125% of the normal operating current) Hence, the fuse rating = 15A \(\times\) (\(\frac{125}{100}\)) = 18.75amps Therefore, the best fuse is 20A.

Shunt resistance is usually given as R = igrg/(I - ig) Where R = 0.05Ω rg = 5Ω I = 2M ∴ ig = IR/(rg + R) = (2 x 0.05)/(5 + 0.05) = 0.10/5.05 = 0.0198A = 19.8mA = 20mA

No expanded working has been attached for this question yet. The verified answer is Option A.

Inductive reactance XL = 2πfL ∴ 2πfL = 2000Ω then f = 2000Ω/2 x π x 10 f = 100/πHz

No expanded working has been attached for this question yet. The verified answer is Option D.

Gamma rays are electromagnetic radiation of nuclear origin; they are produced from energy changes which occur within the nucleus of atoms

Energy of a particle is given by E = hf = hV/λ (since V = fxλ) But in terms of mass M, E = ∆MV2 ∴ ∆MV2 = hV/λ ∆MV = h/λ ∴ V = h/λ∆M V = 6.63 x 10-34/(10-27 x 10-8) V = 6.63 x 10-34/10-35 V = 6.63 x 10-34+35 V = 6.63 x 10 V = 66.3ms-1