Maths

Two cells of emf $\varepsilon_{1}$ and $\varepsilon_{2}$, internal resistance $r_{1}$ and $r_{2}$, connected in parallel. The equivalent emf of the combination is- (A) $\frac{\varepsilon_{1} r_{1}+\varepsilon_{1} r_{1}}{r_{1}+r_{2}}$ (B) $\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$ (C) $\sqrt{\varepsilon_{1} \times \varepsilon_{2}}$ (D) $\frac{\varepsilon_{1}+\varepsilon_{2}}{2}$

Answer: Option (B)$\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$

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(a) In the figure (i) given below, AB is a diameter of the circle. If ∠ADC = 120°, find ∠CAB. (b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.

Solution: (a) Construction: Join BC, and AC then ABCD is a cyclic quadrilateral. Now in ∆DCF Ext. ∠2 = x + z and in ∆CBE Ext. ∠1 = x + y Adding (i) and (ii) x + y + x + z = ∠1 + ∠2 2 x + y + z =...

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(a) In the figure (i) given below, M, A, B, N are points on a circle having centre O. AN and MB cut at Y. If ∠NYB = 50° and ∠YNB = 20°, find ∠MAN and the reflex angle MON. (b) In the figure (ii) given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find (i) ∠ACB (ii) ∠OBC (iii) ∠OAB (iv) ∠CBA

Solution (a) ∠NYB = 50°, ∠YNB = 20°. In ∆YNB, ∠NYB + ∠YNB + ∠YBN = 180o 50o + 20o + ∠YBN = 180o ∠YBN + 70o = 180o ∠YBN = 180o – 70o = 110o But ∠MAN = ∠YBN (Angles in the same segment) ∠MAN = 110o...

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