
A uniform iron rod of length $22.5$ m weighs $85.5$ kg respectively. How much does the similar rod of length $5$ m weigh?
(a) \[20\]kg
(b) \[19\] kg
(c) Cannot be determined
(d) None of these
Answer
490.2k+ views
Hint: The given problem revolves around the concepts of measurements. As a result, of finding the required value the given question resembles that the iron rod uniform which tremendously denotes that density (or, weight) of the rod is the same throughout the whole rod which might get extended as per requirement (Mathematically expressed as, weight per unit length of the rod $ = \dfrac{{weight{\text{ }}of{\text{ }}the{\text{ }}rod}}{{lenght{\text{ }}of{\text{ }}the{\text{ }}rod}}$ respectively). Hence, substituting the values and then multiplying with respect to the desired measurement, say, $5$ m length, to get the required value.
Complete step-by-step answer:
Since, we have given that to find the weight (in kg) of iron rod which is $22.5$ m in length having the weight of $85.5$ kg respectively
It’s very simple to calculate any length for the same rod having same respective dimensions particularly,
As a result, to calculate the weight of iron rod of $5$ m in length
Hence, since by assuming the condition for $1$ m particularly,
Dividing the given weight by length of the respective iron rod that is mathematically expressed as the finding of weight per unit length of iron rod is,
$ = \dfrac{{85.5kg}}{{22.5m}}$
As a result, (since for the further calculations) reducing the above equation in simplest form i.e. by multiplying and dividing by $5$, we get
$ = \dfrac{{17.1kg}}{{4.5m}}$ … (i)
Since, we have given that the iron rod is “uniform” which seems that it may extended with same condition,
As a result, equation (i) satisfies by multiplying it by $5$ m length (to get the new weight of the respective rod), we get
$ = \dfrac{{17.1kg}}{{4.5m}} \times 5m$
Solving the equation mathematically, we get
$ = \dfrac{{17.1kg}}{{0.9}} = \dfrac{{171}}{9}kg$
$ = 19kg$
Hence, similar rod of length $5$ m weighs $19$ kg respectively.
So, the correct answer is “Option b”.
Note: No need to think in a complex way. The problem revolves simply by dividing weight by length i.e. $ = \dfrac{{weight{\text{ }}of{\text{ }}the{\text{ }}rod}}{{lenght{\text{ }}of{\text{ }}the{\text{ }}rod}}$ and then multiplying with desired parameter. Remember, (in this case particularly) solve the required value by considering the unit parameters that are always, so as to be sure of our final answer.
Complete step-by-step answer:
Since, we have given that to find the weight (in kg) of iron rod which is $22.5$ m in length having the weight of $85.5$ kg respectively
It’s very simple to calculate any length for the same rod having same respective dimensions particularly,
As a result, to calculate the weight of iron rod of $5$ m in length
Hence, since by assuming the condition for $1$ m particularly,
Dividing the given weight by length of the respective iron rod that is mathematically expressed as the finding of weight per unit length of iron rod is,
$ = \dfrac{{85.5kg}}{{22.5m}}$
As a result, (since for the further calculations) reducing the above equation in simplest form i.e. by multiplying and dividing by $5$, we get
$ = \dfrac{{17.1kg}}{{4.5m}}$ … (i)
Since, we have given that the iron rod is “uniform” which seems that it may extended with same condition,
As a result, equation (i) satisfies by multiplying it by $5$ m length (to get the new weight of the respective rod), we get
$ = \dfrac{{17.1kg}}{{4.5m}} \times 5m$
Solving the equation mathematically, we get
$ = \dfrac{{17.1kg}}{{0.9}} = \dfrac{{171}}{9}kg$
$ = 19kg$
Hence, similar rod of length $5$ m weighs $19$ kg respectively.
So, the correct answer is “Option b”.
Note: No need to think in a complex way. The problem revolves simply by dividing weight by length i.e. $ = \dfrac{{weight{\text{ }}of{\text{ }}the{\text{ }}rod}}{{lenght{\text{ }}of{\text{ }}the{\text{ }}rod}}$ and then multiplying with desired parameter. Remember, (in this case particularly) solve the required value by considering the unit parameters that are always, so as to be sure of our final answer.
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