139924
The diameter and height of a cylinder are measured by a meter scale to be \(12.6 \pm 0.1 \mathrm{~cm}\) and \(34.2 \pm 0.1 \mathrm{~cm}\), respectively. What will be the value of its volume in appropriate significant figures?
139925
The area of a square is \(5.29 \mathrm{~cm}^{2}\). The area of 7 such squares taking into account the significant figures is
1 \(37.030 \mathrm{~cm}^{2}\)
2 \(37.0 \mathrm{~cm}^{2}\)
3 \(37.03 \mathrm{~cm}^{2}\)
4 \(37 \mathrm{~cm}^{2}\)
Explanation:
B Given, Area of a square \((\mathrm{A})=5.29 \mathrm{~cm}^{2}\) Total area \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\ldots \ldots \mathrm{A}_{7}\) \(=\mathrm{A}+\mathrm{A}+\ldots . .7\) times \(=7 \mathrm{~A}=7 \times 5.29=37.03 \mathrm{~cm}^{2}\) The result should have three signification figure. Hence, \(\mathrm{A}=37.0 \mathrm{~cm}^{2}\)
JEE Main-09.04.2019
Units and Measurements
139926
Identify the incorrect statement among the following.
1 A true length of \(5.678 \mathrm{~km}\) has been measured in two experiments as \(5.5 \mathrm{~km}\) and \(5.51 \mathrm{~km}\), respectively. The second measurement has more precision.
2 Length of \(1 \mathrm{~m}\) and \(0.5 \mathrm{~m}\) have been both measured with the same absolute error of 0.01 \(\mathrm{m}\). Both the measurement are equally accurate.
3 The numbers of significant digits in 1.6 and 0.60 are both two.
4 The number 2.445 can be rounded to two decimal place as 2.45 .
Explanation:
B , d) : As the second measured value i.e, \(5.51 \mathrm{~km}\) is more closed to the result i.e, \(5.678 \mathrm{~km}\). So, it is more precise. Hence, the statement (a) is correct. The percentage error of in case of \(1 \mathrm{~m}\) length is- \(\%\) Error in \(1 \mathrm{~m}\) length \(=\frac{\Delta l_{1}}{l_{1}} \times 100=\frac{0.01}{1} \times 100=1 \%\) \(\%\) Error in \(0.5 \mathrm{~m}\) length \(=\frac{\Delta l_{2}}{l_{2}} \times 100=\frac{0.01}{0.05} \times 100=2 \%\) Hence, the statement (b) is incorrect. No. of significant in \(1.6=(1,6)=2\) significant figure And no. of significant in \(0.60=(6,0)=2\) significant figure Hence, the statement (c) is correct. The number 2.445 can be rounded to two decimal place as 2.44 . Hence, the statement (d) is incorrect. So, option (b) and (d) will be incorrect.
TS- EAMCET-03.05.2019
Units and Measurements
139928
Assertion (A) : The number 0.00764 has three significant figures. Reason (R) : If the number is less than 1 , the zeros on the right of the decimal point but to the left of the first non-zero digit are not significant.
1 Both (A) and (R) are true and (R) is the correct explanation of \((\mathrm{A})\).
2 Both (A) and (R) are true but (R) is not the correct explanation of \((\mathrm{A})\).
3 (A) is true but (R) is false.
4 (A) is false but (R) is true.
Explanation:
A For Assertion (A):- No. of significant figure in 0.00764 \(=(7,6,4)\) There are 3 significant figure in above number For reason \((\mathrm{R})\) :- Let a number is \(0.00 \text { abc } 00 \rightarrow \text { significant (Right zeros) }\) \(\downarrow\) \(\text { Non - significant(Left zeros) }\)
AP EAMCET (23.04.2019) Shift-I
Units and Measurements
139929
The value of resistance is \(10.845 \Omega\) and the current is \(3.23 \mathrm{~A}\). On multiplying, we get the potential difference is \(35.02935 \mathrm{~V}\). The value of potential difference in terms of significant figures would be
1 \(35 \mathrm{~V}\)
2 \(35.0 \mathrm{~V}\)
3 \(35.029 \mathrm{~V}\)
4 \(35.03 \mathrm{~V}\)
Explanation:
B Resistance \((\mathrm{R})=10.845 \Omega\) Current \((\mathrm{I})=3.23 \mathrm{~A}\) \(\mathrm{V}=\mathrm{IR}\) (By Ohm's law) Where, \(\mathrm{V}=\) Voltage, \(\mathrm{I}=\) Current, \(\mathrm{R}=\) Resistance \(\mathrm{V}=10.845 \times 3.23=35.02935 \mathrm{~V}\) Rounding of the value of potential difference to retain their significant figure in result \(=35.0\) volt . Since, (1). If it less than 5 , drop it and all the figure to the right of it. (2). If it is more than 5 , increased by 1 the number to be rounded, that is the proceeding figure. Hence, 2 is less than 5 , then the proceeding figure is same.
139924
The diameter and height of a cylinder are measured by a meter scale to be \(12.6 \pm 0.1 \mathrm{~cm}\) and \(34.2 \pm 0.1 \mathrm{~cm}\), respectively. What will be the value of its volume in appropriate significant figures?
139925
The area of a square is \(5.29 \mathrm{~cm}^{2}\). The area of 7 such squares taking into account the significant figures is
1 \(37.030 \mathrm{~cm}^{2}\)
2 \(37.0 \mathrm{~cm}^{2}\)
3 \(37.03 \mathrm{~cm}^{2}\)
4 \(37 \mathrm{~cm}^{2}\)
Explanation:
B Given, Area of a square \((\mathrm{A})=5.29 \mathrm{~cm}^{2}\) Total area \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\ldots \ldots \mathrm{A}_{7}\) \(=\mathrm{A}+\mathrm{A}+\ldots . .7\) times \(=7 \mathrm{~A}=7 \times 5.29=37.03 \mathrm{~cm}^{2}\) The result should have three signification figure. Hence, \(\mathrm{A}=37.0 \mathrm{~cm}^{2}\)
JEE Main-09.04.2019
Units and Measurements
139926
Identify the incorrect statement among the following.
1 A true length of \(5.678 \mathrm{~km}\) has been measured in two experiments as \(5.5 \mathrm{~km}\) and \(5.51 \mathrm{~km}\), respectively. The second measurement has more precision.
2 Length of \(1 \mathrm{~m}\) and \(0.5 \mathrm{~m}\) have been both measured with the same absolute error of 0.01 \(\mathrm{m}\). Both the measurement are equally accurate.
3 The numbers of significant digits in 1.6 and 0.60 are both two.
4 The number 2.445 can be rounded to two decimal place as 2.45 .
Explanation:
B , d) : As the second measured value i.e, \(5.51 \mathrm{~km}\) is more closed to the result i.e, \(5.678 \mathrm{~km}\). So, it is more precise. Hence, the statement (a) is correct. The percentage error of in case of \(1 \mathrm{~m}\) length is- \(\%\) Error in \(1 \mathrm{~m}\) length \(=\frac{\Delta l_{1}}{l_{1}} \times 100=\frac{0.01}{1} \times 100=1 \%\) \(\%\) Error in \(0.5 \mathrm{~m}\) length \(=\frac{\Delta l_{2}}{l_{2}} \times 100=\frac{0.01}{0.05} \times 100=2 \%\) Hence, the statement (b) is incorrect. No. of significant in \(1.6=(1,6)=2\) significant figure And no. of significant in \(0.60=(6,0)=2\) significant figure Hence, the statement (c) is correct. The number 2.445 can be rounded to two decimal place as 2.44 . Hence, the statement (d) is incorrect. So, option (b) and (d) will be incorrect.
TS- EAMCET-03.05.2019
Units and Measurements
139928
Assertion (A) : The number 0.00764 has three significant figures. Reason (R) : If the number is less than 1 , the zeros on the right of the decimal point but to the left of the first non-zero digit are not significant.
1 Both (A) and (R) are true and (R) is the correct explanation of \((\mathrm{A})\).
2 Both (A) and (R) are true but (R) is not the correct explanation of \((\mathrm{A})\).
3 (A) is true but (R) is false.
4 (A) is false but (R) is true.
Explanation:
A For Assertion (A):- No. of significant figure in 0.00764 \(=(7,6,4)\) There are 3 significant figure in above number For reason \((\mathrm{R})\) :- Let a number is \(0.00 \text { abc } 00 \rightarrow \text { significant (Right zeros) }\) \(\downarrow\) \(\text { Non - significant(Left zeros) }\)
AP EAMCET (23.04.2019) Shift-I
Units and Measurements
139929
The value of resistance is \(10.845 \Omega\) and the current is \(3.23 \mathrm{~A}\). On multiplying, we get the potential difference is \(35.02935 \mathrm{~V}\). The value of potential difference in terms of significant figures would be
1 \(35 \mathrm{~V}\)
2 \(35.0 \mathrm{~V}\)
3 \(35.029 \mathrm{~V}\)
4 \(35.03 \mathrm{~V}\)
Explanation:
B Resistance \((\mathrm{R})=10.845 \Omega\) Current \((\mathrm{I})=3.23 \mathrm{~A}\) \(\mathrm{V}=\mathrm{IR}\) (By Ohm's law) Where, \(\mathrm{V}=\) Voltage, \(\mathrm{I}=\) Current, \(\mathrm{R}=\) Resistance \(\mathrm{V}=10.845 \times 3.23=35.02935 \mathrm{~V}\) Rounding of the value of potential difference to retain their significant figure in result \(=35.0\) volt . Since, (1). If it less than 5 , drop it and all the figure to the right of it. (2). If it is more than 5 , increased by 1 the number to be rounded, that is the proceeding figure. Hence, 2 is less than 5 , then the proceeding figure is same.
139924
The diameter and height of a cylinder are measured by a meter scale to be \(12.6 \pm 0.1 \mathrm{~cm}\) and \(34.2 \pm 0.1 \mathrm{~cm}\), respectively. What will be the value of its volume in appropriate significant figures?
139925
The area of a square is \(5.29 \mathrm{~cm}^{2}\). The area of 7 such squares taking into account the significant figures is
1 \(37.030 \mathrm{~cm}^{2}\)
2 \(37.0 \mathrm{~cm}^{2}\)
3 \(37.03 \mathrm{~cm}^{2}\)
4 \(37 \mathrm{~cm}^{2}\)
Explanation:
B Given, Area of a square \((\mathrm{A})=5.29 \mathrm{~cm}^{2}\) Total area \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\ldots \ldots \mathrm{A}_{7}\) \(=\mathrm{A}+\mathrm{A}+\ldots . .7\) times \(=7 \mathrm{~A}=7 \times 5.29=37.03 \mathrm{~cm}^{2}\) The result should have three signification figure. Hence, \(\mathrm{A}=37.0 \mathrm{~cm}^{2}\)
JEE Main-09.04.2019
Units and Measurements
139926
Identify the incorrect statement among the following.
1 A true length of \(5.678 \mathrm{~km}\) has been measured in two experiments as \(5.5 \mathrm{~km}\) and \(5.51 \mathrm{~km}\), respectively. The second measurement has more precision.
2 Length of \(1 \mathrm{~m}\) and \(0.5 \mathrm{~m}\) have been both measured with the same absolute error of 0.01 \(\mathrm{m}\). Both the measurement are equally accurate.
3 The numbers of significant digits in 1.6 and 0.60 are both two.
4 The number 2.445 can be rounded to two decimal place as 2.45 .
Explanation:
B , d) : As the second measured value i.e, \(5.51 \mathrm{~km}\) is more closed to the result i.e, \(5.678 \mathrm{~km}\). So, it is more precise. Hence, the statement (a) is correct. The percentage error of in case of \(1 \mathrm{~m}\) length is- \(\%\) Error in \(1 \mathrm{~m}\) length \(=\frac{\Delta l_{1}}{l_{1}} \times 100=\frac{0.01}{1} \times 100=1 \%\) \(\%\) Error in \(0.5 \mathrm{~m}\) length \(=\frac{\Delta l_{2}}{l_{2}} \times 100=\frac{0.01}{0.05} \times 100=2 \%\) Hence, the statement (b) is incorrect. No. of significant in \(1.6=(1,6)=2\) significant figure And no. of significant in \(0.60=(6,0)=2\) significant figure Hence, the statement (c) is correct. The number 2.445 can be rounded to two decimal place as 2.44 . Hence, the statement (d) is incorrect. So, option (b) and (d) will be incorrect.
TS- EAMCET-03.05.2019
Units and Measurements
139928
Assertion (A) : The number 0.00764 has three significant figures. Reason (R) : If the number is less than 1 , the zeros on the right of the decimal point but to the left of the first non-zero digit are not significant.
1 Both (A) and (R) are true and (R) is the correct explanation of \((\mathrm{A})\).
2 Both (A) and (R) are true but (R) is not the correct explanation of \((\mathrm{A})\).
3 (A) is true but (R) is false.
4 (A) is false but (R) is true.
Explanation:
A For Assertion (A):- No. of significant figure in 0.00764 \(=(7,6,4)\) There are 3 significant figure in above number For reason \((\mathrm{R})\) :- Let a number is \(0.00 \text { abc } 00 \rightarrow \text { significant (Right zeros) }\) \(\downarrow\) \(\text { Non - significant(Left zeros) }\)
AP EAMCET (23.04.2019) Shift-I
Units and Measurements
139929
The value of resistance is \(10.845 \Omega\) and the current is \(3.23 \mathrm{~A}\). On multiplying, we get the potential difference is \(35.02935 \mathrm{~V}\). The value of potential difference in terms of significant figures would be
1 \(35 \mathrm{~V}\)
2 \(35.0 \mathrm{~V}\)
3 \(35.029 \mathrm{~V}\)
4 \(35.03 \mathrm{~V}\)
Explanation:
B Resistance \((\mathrm{R})=10.845 \Omega\) Current \((\mathrm{I})=3.23 \mathrm{~A}\) \(\mathrm{V}=\mathrm{IR}\) (By Ohm's law) Where, \(\mathrm{V}=\) Voltage, \(\mathrm{I}=\) Current, \(\mathrm{R}=\) Resistance \(\mathrm{V}=10.845 \times 3.23=35.02935 \mathrm{~V}\) Rounding of the value of potential difference to retain their significant figure in result \(=35.0\) volt . Since, (1). If it less than 5 , drop it and all the figure to the right of it. (2). If it is more than 5 , increased by 1 the number to be rounded, that is the proceeding figure. Hence, 2 is less than 5 , then the proceeding figure is same.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Units and Measurements
139924
The diameter and height of a cylinder are measured by a meter scale to be \(12.6 \pm 0.1 \mathrm{~cm}\) and \(34.2 \pm 0.1 \mathrm{~cm}\), respectively. What will be the value of its volume in appropriate significant figures?
139925
The area of a square is \(5.29 \mathrm{~cm}^{2}\). The area of 7 such squares taking into account the significant figures is
1 \(37.030 \mathrm{~cm}^{2}\)
2 \(37.0 \mathrm{~cm}^{2}\)
3 \(37.03 \mathrm{~cm}^{2}\)
4 \(37 \mathrm{~cm}^{2}\)
Explanation:
B Given, Area of a square \((\mathrm{A})=5.29 \mathrm{~cm}^{2}\) Total area \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\ldots \ldots \mathrm{A}_{7}\) \(=\mathrm{A}+\mathrm{A}+\ldots . .7\) times \(=7 \mathrm{~A}=7 \times 5.29=37.03 \mathrm{~cm}^{2}\) The result should have three signification figure. Hence, \(\mathrm{A}=37.0 \mathrm{~cm}^{2}\)
JEE Main-09.04.2019
Units and Measurements
139926
Identify the incorrect statement among the following.
1 A true length of \(5.678 \mathrm{~km}\) has been measured in two experiments as \(5.5 \mathrm{~km}\) and \(5.51 \mathrm{~km}\), respectively. The second measurement has more precision.
2 Length of \(1 \mathrm{~m}\) and \(0.5 \mathrm{~m}\) have been both measured with the same absolute error of 0.01 \(\mathrm{m}\). Both the measurement are equally accurate.
3 The numbers of significant digits in 1.6 and 0.60 are both two.
4 The number 2.445 can be rounded to two decimal place as 2.45 .
Explanation:
B , d) : As the second measured value i.e, \(5.51 \mathrm{~km}\) is more closed to the result i.e, \(5.678 \mathrm{~km}\). So, it is more precise. Hence, the statement (a) is correct. The percentage error of in case of \(1 \mathrm{~m}\) length is- \(\%\) Error in \(1 \mathrm{~m}\) length \(=\frac{\Delta l_{1}}{l_{1}} \times 100=\frac{0.01}{1} \times 100=1 \%\) \(\%\) Error in \(0.5 \mathrm{~m}\) length \(=\frac{\Delta l_{2}}{l_{2}} \times 100=\frac{0.01}{0.05} \times 100=2 \%\) Hence, the statement (b) is incorrect. No. of significant in \(1.6=(1,6)=2\) significant figure And no. of significant in \(0.60=(6,0)=2\) significant figure Hence, the statement (c) is correct. The number 2.445 can be rounded to two decimal place as 2.44 . Hence, the statement (d) is incorrect. So, option (b) and (d) will be incorrect.
TS- EAMCET-03.05.2019
Units and Measurements
139928
Assertion (A) : The number 0.00764 has three significant figures. Reason (R) : If the number is less than 1 , the zeros on the right of the decimal point but to the left of the first non-zero digit are not significant.
1 Both (A) and (R) are true and (R) is the correct explanation of \((\mathrm{A})\).
2 Both (A) and (R) are true but (R) is not the correct explanation of \((\mathrm{A})\).
3 (A) is true but (R) is false.
4 (A) is false but (R) is true.
Explanation:
A For Assertion (A):- No. of significant figure in 0.00764 \(=(7,6,4)\) There are 3 significant figure in above number For reason \((\mathrm{R})\) :- Let a number is \(0.00 \text { abc } 00 \rightarrow \text { significant (Right zeros) }\) \(\downarrow\) \(\text { Non - significant(Left zeros) }\)
AP EAMCET (23.04.2019) Shift-I
Units and Measurements
139929
The value of resistance is \(10.845 \Omega\) and the current is \(3.23 \mathrm{~A}\). On multiplying, we get the potential difference is \(35.02935 \mathrm{~V}\). The value of potential difference in terms of significant figures would be
1 \(35 \mathrm{~V}\)
2 \(35.0 \mathrm{~V}\)
3 \(35.029 \mathrm{~V}\)
4 \(35.03 \mathrm{~V}\)
Explanation:
B Resistance \((\mathrm{R})=10.845 \Omega\) Current \((\mathrm{I})=3.23 \mathrm{~A}\) \(\mathrm{V}=\mathrm{IR}\) (By Ohm's law) Where, \(\mathrm{V}=\) Voltage, \(\mathrm{I}=\) Current, \(\mathrm{R}=\) Resistance \(\mathrm{V}=10.845 \times 3.23=35.02935 \mathrm{~V}\) Rounding of the value of potential difference to retain their significant figure in result \(=35.0\) volt . Since, (1). If it less than 5 , drop it and all the figure to the right of it. (2). If it is more than 5 , increased by 1 the number to be rounded, that is the proceeding figure. Hence, 2 is less than 5 , then the proceeding figure is same.
139924
The diameter and height of a cylinder are measured by a meter scale to be \(12.6 \pm 0.1 \mathrm{~cm}\) and \(34.2 \pm 0.1 \mathrm{~cm}\), respectively. What will be the value of its volume in appropriate significant figures?
139925
The area of a square is \(5.29 \mathrm{~cm}^{2}\). The area of 7 such squares taking into account the significant figures is
1 \(37.030 \mathrm{~cm}^{2}\)
2 \(37.0 \mathrm{~cm}^{2}\)
3 \(37.03 \mathrm{~cm}^{2}\)
4 \(37 \mathrm{~cm}^{2}\)
Explanation:
B Given, Area of a square \((\mathrm{A})=5.29 \mathrm{~cm}^{2}\) Total area \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\ldots \ldots \mathrm{A}_{7}\) \(=\mathrm{A}+\mathrm{A}+\ldots . .7\) times \(=7 \mathrm{~A}=7 \times 5.29=37.03 \mathrm{~cm}^{2}\) The result should have three signification figure. Hence, \(\mathrm{A}=37.0 \mathrm{~cm}^{2}\)
JEE Main-09.04.2019
Units and Measurements
139926
Identify the incorrect statement among the following.
1 A true length of \(5.678 \mathrm{~km}\) has been measured in two experiments as \(5.5 \mathrm{~km}\) and \(5.51 \mathrm{~km}\), respectively. The second measurement has more precision.
2 Length of \(1 \mathrm{~m}\) and \(0.5 \mathrm{~m}\) have been both measured with the same absolute error of 0.01 \(\mathrm{m}\). Both the measurement are equally accurate.
3 The numbers of significant digits in 1.6 and 0.60 are both two.
4 The number 2.445 can be rounded to two decimal place as 2.45 .
Explanation:
B , d) : As the second measured value i.e, \(5.51 \mathrm{~km}\) is more closed to the result i.e, \(5.678 \mathrm{~km}\). So, it is more precise. Hence, the statement (a) is correct. The percentage error of in case of \(1 \mathrm{~m}\) length is- \(\%\) Error in \(1 \mathrm{~m}\) length \(=\frac{\Delta l_{1}}{l_{1}} \times 100=\frac{0.01}{1} \times 100=1 \%\) \(\%\) Error in \(0.5 \mathrm{~m}\) length \(=\frac{\Delta l_{2}}{l_{2}} \times 100=\frac{0.01}{0.05} \times 100=2 \%\) Hence, the statement (b) is incorrect. No. of significant in \(1.6=(1,6)=2\) significant figure And no. of significant in \(0.60=(6,0)=2\) significant figure Hence, the statement (c) is correct. The number 2.445 can be rounded to two decimal place as 2.44 . Hence, the statement (d) is incorrect. So, option (b) and (d) will be incorrect.
TS- EAMCET-03.05.2019
Units and Measurements
139928
Assertion (A) : The number 0.00764 has three significant figures. Reason (R) : If the number is less than 1 , the zeros on the right of the decimal point but to the left of the first non-zero digit are not significant.
1 Both (A) and (R) are true and (R) is the correct explanation of \((\mathrm{A})\).
2 Both (A) and (R) are true but (R) is not the correct explanation of \((\mathrm{A})\).
3 (A) is true but (R) is false.
4 (A) is false but (R) is true.
Explanation:
A For Assertion (A):- No. of significant figure in 0.00764 \(=(7,6,4)\) There are 3 significant figure in above number For reason \((\mathrm{R})\) :- Let a number is \(0.00 \text { abc } 00 \rightarrow \text { significant (Right zeros) }\) \(\downarrow\) \(\text { Non - significant(Left zeros) }\)
AP EAMCET (23.04.2019) Shift-I
Units and Measurements
139929
The value of resistance is \(10.845 \Omega\) and the current is \(3.23 \mathrm{~A}\). On multiplying, we get the potential difference is \(35.02935 \mathrm{~V}\). The value of potential difference in terms of significant figures would be
1 \(35 \mathrm{~V}\)
2 \(35.0 \mathrm{~V}\)
3 \(35.029 \mathrm{~V}\)
4 \(35.03 \mathrm{~V}\)
Explanation:
B Resistance \((\mathrm{R})=10.845 \Omega\) Current \((\mathrm{I})=3.23 \mathrm{~A}\) \(\mathrm{V}=\mathrm{IR}\) (By Ohm's law) Where, \(\mathrm{V}=\) Voltage, \(\mathrm{I}=\) Current, \(\mathrm{R}=\) Resistance \(\mathrm{V}=10.845 \times 3.23=35.02935 \mathrm{~V}\) Rounding of the value of potential difference to retain their significant figure in result \(=35.0\) volt . Since, (1). If it less than 5 , drop it and all the figure to the right of it. (2). If it is more than 5 , increased by 1 the number to be rounded, that is the proceeding figure. Hence, 2 is less than 5 , then the proceeding figure is same.