I'd say b, precise, here.
If there's an error somewhere in the experiment or project, then it is consistently .... wrong. So, just 'cos you measure something precisely, it doesn't mean that you've measured it accurately. Maybe an example would be a measurement of length. If you used a metal ruler at zero degrees C, you can measure to say half a millimetre. A series of measurements of the same object would give very similar readings. But, if you used same metal ruler at, say 100 celsius (implausible) then you'd probably get a different set of readings. 'cos of the expansion of the metal ruler.
Answer:
Solution
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Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
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SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
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A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
They are related because they both have a potential to store energy
Answer: 1.58Ω
Explanation:
R1 = 5Ω
R2 = 15Ω
R3 = 3Ω
R4 = 30Ω
The equivalent resistance of the resistor connection (Rtotal) of the four resistors connected in parallel is obtained by adding the reciprocal of each resistance.
i.e 1/Rtotal = (1/R1 + 1/R2 + 1/R3 +1/R4)
1/Rtotal = (1/5Ω + 1/15Ω + 1/3Ω + 1/30Ω)
Make 30Ω the lowest common multiple
1/Rtotal = (6 + 2 + 10 + 1)/30Ω
1/Rtotal = 19/30Ω
To get the value of Rtotal, cross multiply
1 x 30Ω = Rtotal x 19
30Ω = Rtotal x 19
Rtotal = 30Ω/19
Rtotal = 1.58Ω
Thus, the equivalent resistance of this resistor connection is 1.58Ω