If you want to learn more about the nature and properties of matter and energy or you're simply preparing for a Physics exam, these Physics past questions and answers are ideal for you.
\( 2.93 \times 10^{15}\)Hz
\( 3.59 \times 10^{16}\)Hz
\( 5.39 \times 10^{15}\)Hz
\( 1.24 \times 10^{15}\)Hz
Correct answer is D
h = \( 6.6 \times 10^{-34}\)Js, 1eV = \( 1.6 \times 10^{-19}\)J, \(E_1\) = - 8.6eV, \(E_2\) = - 3.5eV, f = ?
ΔE = hf
f = \(\frac{ΔE}{h}\)
ΔE = \(E_2\) - \(E_1\) = - 3.5eV - (- 8.6eV) = 5.1eV
but 5.1eV = 5.1 X \( 1.6 \times 10^{-19}\) = \( 8.16 \times 10^{-19}\)J
f = \(\frac{8.16 \times 10^{-19}}{ 6.6 \times 10^{-34}}\)
Therefore, f = \( 1.24 \times 10^{15}\)Hz
longer than \(10^{-6} m\).
shorter than \(10^{-6} m\).
shorter than \(10^{3} m\).
longer than \(10^{3} m\).
Correct answer is A
The range of wavelengths for infrared waves is given as \(10^{-3}\) m to \(10^{-6}\)m. This means that the wavelength of the radiation is longer than \(10^{-6} m\) but shorter than \{10^{-3} m\). Therefore, the only correct option is that the wavelength is longer than \(10^{-6} m\).
\(30^0\)
\(60^0\)
\(45^0\)
\(0^0\)
Correct answer is C
Maximum Range R = \(\frac{u^2sin2\theta}{g}\)
maximum height H = \(\frac{u^2sin^2\theta}{2g}\)
But R : H = 4
\(\frac{u^2sin2\theta}{g}\) : \(\frac{u^2sin^2\theta}{2g}\) = 4
\(\frac{u^2sin2\theta}{g} \times\frac{2g}{u^2sin^2\theta}\) = 4
\(\frac{2sin2\theta}{sin^2\theta}\) = 4
since 2sinΘ = 2sinΘcosΘ
\(\frac{2 \times 2sin\theta cos\theta}{sin^2\theta}\) = 4
\(\frac{4sin\theta cos\theta}{sin^2\theta}\) = 4
\(\frac{sin\theta cos\theta}{sin^2\theta}\) = 1
\(\frac{cos\theta}{sin\theta}\) = 1
\(\frac{sin\theta}{cos\theta}\) = 1
\(tan \theta\) = 1
\(\theta = tan^{-1}1\) = 45°
A charge of 10 C is transferred across a potential difference of 220 V. Determine the work done.
22J
2200J
210J
230J
Correct answer is B
The work done in moving a charge across a potential difference is given by the formula W = QV, where W is the work done, Q is the charge and V is the potential difference. Substituting the given values, we get W = 10 C * 220 V = 2200 J. Hence, the work done is 2200 J.
1.550m
0.646m
5.120m
0.195m
Correct answer is B
f = 512Hz, V = 331m/s λ = ?
V = f λ
λ = \(\frac{v}{f} = \frac{331}{512}\)
λ = 0.646m.
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