Question

QUESTION ❓ ✔️❓

1) In a workshop a worker measures the length of a steel plate with a Vernier callipers having a least count 0.01 cm. Four such measurements of the length yielded the following values:
3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm.
Find the mean length, the mean absolute error and the percentage error in the measured value of the length.

QUESTION ❓✔️❓



2)The following observations were taken for determining surface tension T of water by capillary method:

diameter of capillary,
D= 1.25 x 10-²m,
rise of water, h = 1.45 x 10-² m.

Using g = 9.80 m/s² and the simplified relation

T = rhg /2 = x 10³ N/m,

The possible error in surface tension is closest to: (JEE (Main) 2017)

(A) 0.15% (B) 1.5% (C) 2.4 % (D) 10%




QUESTION ❓✔️❓



3) A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm.
The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level.

If screw gauge has a zero error of
- 0.004 cm, the correct diameter of the ball is (NEET (UG) 2018

(A) 0.521 cm (B) 0.525 cm

(C) 0.053 cm (D) 0.529 cm​

Answer 1

$\large \rm {\underbrace{\underline{Question\: 1:}}}$

$\sf \red {\underline{\underline{Provided\: that:}}}$

$\sf \to {Least\: count=0.01cm}$

➻Measurements of the length,

$\sf \to {L_{1}=3.11cm}$

$\sf \to {L_{2}=3.13cm}$

$\sf \to {L_{3}=3.14cm}$

$\sf \to {L_{4}=3.14cm}$

$\sf \blue {\underline{\underline{To\: determine:}}}$

$\sf \to {i.)Mean\:length(L_{mean})=?}$

$\sf \to {ii.)Mean\: absolute\: error(∆L_{mean})=?}$

$\sf \to {iii.)Percentage\: Error=?}$

$\sf \pink {\underline{\underline{i.)Mean\:length(L_{mean}):}}}$

$\sf {L_{mean}=\frac{Sum\: of\: all\: the\: measured\: lengths}{No\: of\: measurements}}$

$\sf \to {\frac{L_{1}+L_{2}+L_{3}+L_{4}}{4}}$

$\sf \to {\frac{3.11+3.13+3.14+3.14}{4}}$

$\sf \to {\frac{12.52}{4}}$

$\sf \implies \tt \green {\boxed{\underline{L_{mean}=3.13cm}}}$

$\sf \purple {\underline{\underline{ii.)Mean\: absolute\: error(∆L_{mean}):}}}$

$\sf {∆L_mean=\frac{Mean\: of\: all\: the\: measured\: lengths}{No\: of\: measurements}}$

$\sf {∆L_mean=\frac{∆L_{1}+∆L_{2}+∆L_{3}+∆L_{4}}{4}}$

➻We need to find the mean of all the measurements.

$\sf {∆L=L_{mean}-measured\: value}$

$\sf \to {∆L_{1}=3.13-3.11}$

$\sf \implies {∆L_{1}=0.02cm}$

$\sf \to {∆L_{2}=3.13-3.13}$

$\sf \implies {∆L_{2}=0cm}$

$\sf \to {∆L_{3}=3.13-3.14}$

$\sf \implies {∆L_{3}=0.01cm}$

$\sf \to {∆L_{4}=3.13-3.14}$

$\sf \implies \tt {∆L_{4}=0.01cm}$

➻Supplanting in the formula,

$\sf \to {∆L_mean=\frac{0.02+0+0.01+0.01}{4}}$

$\sf \to \tt {∆L_mean=\frac{0.04}{4}}$

$\sf \implies \green{\boxed{\underline{∆L_{mean}=0.01cm}}}$

$\sf \orange {\underline{\underline{iii.) Percentage\: error:}}}$

$\sf {Percentage\: error=\frac{Mean\: absolute\: error}{Mean\: length}×100}$

$\sf \to{Percentage\: error=\frac{∆L_{mean}}{L_{mean}}×100}$

$\sf \to {Percentage\: error=\frac{0.01cm}{3.13cm}×100}$

$\sf \to {Percentage\: error=\frac{100}{313}}$

$\sf \to {Percentage\: error≈0.319}$

★ᴘᴇʀᴄᴇɴᴛᴀɢᴇ ᴇʀʀᴏʀ=0.32%

━═━═━═━═━═━═━═━═━═━

$\large \rm {\underbrace{\underline{Question\: 2:}}}$

$\sf \red {\underline{\underline{Provided\: that:}}}$

$\sf \to {Diameter(D)= 1.25 x 10-²m}$

$\sf \to {Height(h)= 1.45 x 10-²m}$

$\sf \to {g=9.80 m/s²}$

$\sf \to {T=\frac{rgh}{2}×10^{3}N/m}$

$\sf \to {T=\frac{dgh}{4}×10^{3}N/m}$

$\sf {[∵r=\frac{d}{2}]}$

$\sf \blue {\underline{\underline{To\: determine:}}}$

➻Possible error in surface tension, i.e,

$\sf \to {\frac{∆T}{T}×100=?}$

$\sf \pink {\underline{\underline{We\: know:}}}$

➻By applying logarithm and differentiating the relation of T we get,

$\to \sf {\frac{∆T}{T}×100=(\frac{∆D}{D}+\frac{∆h}{h})×100}$

➻The maximum error in ∆D and ∆H can be written as 0.01

➻Supplanting the given values,

$\to \sf {\frac{∆T}{T}×100=(\frac{0.01}{1.25×10-²}+\frac{0.01}{1.45×10-²})×100}$

$\to \sf {\frac{∆T}{T}×100=(\frac{100}{125 }+\frac{100}{145})×100}$

$\to \sf {\frac{∆T}{T}×100=(0.8+0.68)×100}$

$\to \sf {\frac{∆T}{T}×100=(1.48)×100}$

$\to \sf {\frac{∆T}{T}×100=(1.5)×100}$

★ᴘᴏssɪʙʟᴇ ᴇʀʀᴏʀ ɪɴ sᴜʀғᴀᴄᴇ ᴛᴇɴsɪᴏɴ =1.5%

━═━═━═━═━═━═━═━═━═━

$\large \rm {\underbrace{\underline{Question\: 3:}}}$

$\sf \red {\underline{\underline{Provided\: that:}}}$

$\sf \to {Least\: count=0.001cm}$

$\sf \to {Main\: scale\: reading(M.S.R)=5mm}$

$\sf \to {M.S.R=0.5cm}$

$\sf \to {Vernier\: scale\: divisions(V.S.D)=25}$

$\sf \to {Zero\: error=0.004cm}$

$\sf \blue {\underline{\underline{To\: determine:}}}$

➻Correct diamter of the ball,i.e,

➻Actual reading=?

$\sf \orange {\underline{\underline{We\: know:}}}$

$\to \sf {Observed\: reading=M.S.R+V.S.D×Least\: count}$

$\to \sf {Observed\: reading=0.5+25×0.001}$

$\to \sf {Observed\: reading=0.5+25×0.001}$

$\to \sf {Observed\: reading=0.5+0.025}$

$\to \sf {Observed\: reading=0.525}$

$\to \sf {Actual\: reading=Observed\: reading+zero\: error}$

$\to \sf {Actual\: reading=0.525+0.004}$

$\implies \sf \green {\fbox{\underline{Actual\: reading=0.529cm}}}$

Answer 2

Answer:

The possible error in surface tension is closest to: (JEE (Main) 2017)

(A) 0.15% (B) 1.5% (C) 2.4 % (D) 10%

QUESTION ❓✔️❓

3) A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm.

The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level.

If screw gauge has a zero error of

- 0.004 cm, the correct diameter of the ball is (NEET (UG) 2018

(A) 0.521 cm (B) 0.525 cm

(C) 0.053 cm (D) 0.529 cm