Evaluate \(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\)
0.2
2
100
409
490
Correct answer is B
\(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\)
= \(\log_{10} (\frac{25 \times 32}{8})\)
= \(\log_{10} 100 \)
= 2
Calculate the resistance of the filament of a lamp rates 240 V, 40 W
240Ω
360Ω
720Ω
1440Ω
2880Ω
Correct answer is D
R = V2/P = 2402/40 = 1440Ω
0.0009Ω/π
100Ω
1030Ω
400Ω
2500Ω
Correct answer is D
Given Data: Frequency (F) = \(\frac{500}{π}\) , Inductance (L) = 0.9H, Capacitance (C) = \(2 \times 10^{-6}\)
Total circuit reactance = Inductive reactance ( X\(_L\) ) - Capacitive reactance ( X\(_C\) )
when ( X\(_L\) ) > ( X\(_C\) )
Inductive reactance ( X\(_L\) ) = 2πFL = 2 \(\times\) π \(\times\) \(\frac{500}{π}\) \(\times\) 0.9 = 900Ω
Capacitive reactance ( X\(_C\) ) = \(\frac{1}{2πFC}\) = \(\frac{1}{2 \times π \times 500/π \times 2 \times 10^{-6}}\)
= \(\frac{1}{2 \times 10^{-3}}\) = \(\frac{1}{0.002}\)
= 500Ω
Total circuit reactance = ( X\(_L\) ) - ( X\(_C\) ) = (900 - 500)Ω
=400Ω
Simplify: \((\frac{16}{81})^{\frac{1}{4}}\)
8/27
1/3
4/9
2/3
-4/3
Correct answer is D
\((\frac{16}{81})^{\frac{1}{4}}\)
= \(((\frac{2}{3})^{4})^{\frac{1}{4}}\)
= \(\frac{2}{3}\)
2/3, 0, 6
3/2, 3, 6,
2/3, 3, 8/3
2/3, 3/4, 6
2/3, 3, 1/3
Correct answer is B
\(T_n = 3. 2^{n - 2} \)
\(T_{1} = 3. 2^{1 - 2} = 3. 2^{-1} \)
\(T_1 = \frac{3}{2} \)
\(T_2 = 3. 2^{2 - 2} \)
\(T_2 = 3. 2^0 = 3\)
\(T_3 = 3. 2^{3 - 2} = 3. 2^1 \)
\(T_3 = 6\)
The first 3 terms of the sequence are \(\frac{3}{2}\), 3 and 6.