Choose the appropriate items from the following
Nichrome, Tungsten, Fuse Wire, Nitrogen
a. Which is used as a heating coil ?
b. Which is an alloy of tin and lead?
c. Which is used as filament ?
Answers
Answer:
c)
is there correct answer
Explanation:
Hint: We solve this problem by using the formula of simple interest.
The formula for the simple interest is given as
SI=P×T×R100
Where P
is the principal amount, T
is the time period and R
is the rate of interest per annum.
After finding the simple interest for all years from 1 to 6 we create a table of ordered pairs containing the number of years as x coordinate and simple interest as y co – ordinate so that we can plot them on a graph of a certain scale.
Complete step by step answer:
We are given that the principal amount as
⇒P=8000
We are also given that the rate of interest per annum as
⇒R=5
Let us assume that the time period as T
We know that the formula for the simple interest is given as
SI=P×T×R100
Where, P
is the principal amount, T
is the time period and R
is the rate of interest per annum.
By using the above formula we get the simple interest in terms of time period as
⇒SI=8000×5×T100⇒SI=400T
Now, let us take the time period as 1 year then we get the simple interest of 1 year as
⇒SI1=400×1⇒SI1=400
Now, let us take the time period as 2 years then we get the simple interest for 2 years as
⇒SI2=400×2⇒SI2=800
Now, let us take the time period as 3 years then we get the simple interest for 3 years as
⇒SI3=400×3⇒SI3=1200
Now, let us take the time period as 4 years then we get the simple interest for 4 years as
⇒SI4=400×4⇒SI4=1600
Now, let us take the time period as 5 years then we get the simple interest for 5 years as
⇒SI5=400×5⇒SI5=2000
Now, let us take the time period as 2 years then we get the simple interest for 2 years as
⇒SI6=400×6⇒SI6=2400
Now, let us create a table containing the order pairs containing number of years as x co – ordinate and simple interest as y co – ordinate as follows
Time period (T)
Simple interest (SIi)
Ordered pair(T,SIi)
1 400 (1,400)
2 800 (2,800)
3 1200 (3,1200)
4 1600 (4,1600)
5 2000 (5,2000)
6 2400 (6,2400)
Now, let us consider a scale of the required graph as follows
On X – axis 1 unit equal to 1 year
On Y – axis 1 unit equal to RS. 400
By using the above scale and the coordinate of points in the table we get the graph as
Note:
Students may make mistakes in finding the simple interest of 2, 3, 4, …, 6 years.
By using the simple interest formula for 2 years we get
⇒SI2=8000×5×2100⇒SI2=800
But students may make mistakes in taking the principal value for the time period of 2 years.
The amount after the 1 year is given as
⇒A=8000+400=8400
Answer:
1.Nichrome
2.Fuse wire
3.Tungsten