31.In a common base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor (β) will be A) 49 B) 50 C) 51 D) 48 View Answer Report DiscussAnswer: Option AExplanation:Ic = 5.488mA Ie = 5.6mA α=IcIe = 5.4885.6 β=α(1−α) = 49
32.The half-life period and the mean life period of a radioactive element are denoted respectively by Th and Tm. Then A) Th=Tm B) Th>Tm C) Th<Tm D) Th≥Tm View Answer Report DiscussAnswer: Option CExplanation:Half-life Th = 0.693λ,Tm=1λ Clearly, Th<Tm
33.The equivalent capacitance of the combination of the capacitors is A) 3.20µF B) 7.80µF C) 3.90µF D) 2.16µF View Answer Report DiscussAnswer: Option AExplanation:Equivalent capacitance of two parallel capacitors 10 μF and 6 μF = (10 + 6)μF = 16 μF . This 16 μF capacitor is in series combination with 4μF capacitor, .'. Equivalent capacitance of the entire combination = 16×416+4=6420 = 3.20μF
34.A point charge q is rotated along a circle in the electric field generated by another point charge Q The work done by the electric field on the rotating charge in one complete revolution is A) Zero B) Positive C) Negative D) zero if the charge Q is at the centre and non zero otherwise. View Answer Report DiscussAnswer: Option AExplanation:Work done in the displacement of charge, W=q△V where ΔV : the potential difference between the displacement of charge Hence, work done in complete rotation is zero
35.Which of the following gates will have an output of I? A) D B) A C) B D) C View Answer Report DiscussAnswer: Option DExplanation:(A) is a NAND gate so the output is ¯1×1=¯1=0 (B) is a NOR gate so output is ¯0+1=¯1=0 (C) is a NAND gate so output is ¯0×1=¯0=1 (D) is a XOR gate so output is 0 ⊕ 0 = 0
