29- If we take 200 gms of water at 100°C in a closed beaker and supply 6 X 10^4 Joules of energy to it,
what happens?
(a) There occurs an increase in temperature from 100°C
(b) There occurs a decrease in temperature from 100°C
(c) No change in temperature occurs but state change takes place
(d) No change in temperature of state occurs
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Answer:
512.50 (D)
Step-by-step explanation:
☆Given:
The difference between compound intrest and
Simple intrest is 12.5
☆To find
Principle and compound intrest
☆Solution:
▪P = ?
▪I = 5%
▪N = 2
P[(1+i)^n-1]P[(1+i)n−1] - p × i × t = 12.5
P[(1+0.05)^2-1]P[(1+0.05)2−1] - p × (⁵/₁₀₀) × 2 = 12.5
[(1.05)^2-1] = 0.1025[(1.05)2−1]=0.1025
0.1025p - \frac{10p}{100}10010p = 12.5
\frac{10.25p-10p}{100}10010.25p−10p = 12.5
0.25p = 1250
P= 1250 ÷ .25 = 5,000
CI = P[(1+i)^n-1]CI=P[(1+i)n−1]
CI = 5000 [(1.05)^2-1]
CI = 5000 × 0.1025
CI = 512.50
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