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Chapter 14: Problem 80
The resistance of a wire is \(10 \Omega\). Its length is increased by \(10 \%\) by stretching. The new resistance will now be nearly (A) \(12 \Omega\) (B) \(1.2 \Omega\) (C) \(13 \Omega\) (D) \(11 \Omega\)
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A galvanometer of resistance \(19.5 \Omega\) gives full-scale deflection when a current of \(0.5 \mathrm{~A}\) is passed through it. It is desired to convert it into an ammeter of full-scale current \(20 \mathrm{~A}\). Value of shunt is (A) \(0.5 \Omega\) (B) \(1 \Omega\) (C) \(1.5 \Omega\) (D) \(2 \Omega\)
An ammeter is obtained by shunting a \(30 \Omega\) galvanometer with a \(30 \Omega\) resistance. What additional shunt should be connected across it to double the range? (A) \(15 \Omega\) (B) \(10 \Omega\) (C) \(5 \Omega\) (D) None of these
A potentiometer is a device used for measuring EMF and internal resistance of a cell. It consists of two circuits, one is main circuit in which there is a cell of given emf \(\varepsilon^{\prime}\) and given resistance \(R\) which is connected across a wire of length \(100 \mathrm{~cm}\) and having resistance \(r\) and another circuit having unknown EMF \(\varepsilon\) and galvanometer. For a given potentiometer, if \(\varepsilon^{\prime}=30 \mathrm{~V}, r=1 \Omega\), and resistance \(R\) varies with time \(t\) given by \(R=2 t\). The jockey can move on wire with constant velocity \(10 \mathrm{~cm} / \mathrm{s}\) and switch \(S\) is closed at \(t=0\) If balancing length is found to be \(70 \mathrm{~cm}\), then the time after which jockey starts moving from \(A\) is (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)
The same mass of copper is drawn into two wires \(1 \mathrm{~mm}\) and \(2 \mathrm{~mm}\) thick. Two wires are connected in series and current is passed through them. Heat produced in the wire is in the ratio (A) \(2: 1\) (B) \(1: 16\) (C) \(4: 1\) (D) \(16: 1\)
The resistance of the series combination of two resistances is \(S\). When they are joined in parallel, the total resistance is \(P\). If \(S=n P\), then the minimum possible value of \(n\) is (A) 4 (B) 3 (C) 2 (D) 1
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