Class 12th

Density of nucleus is constant.

R = R? [1 + α? T]
⇒ 16 = R? [1 + α (15 - T? )]
20 = R? [1 + α (100 - T? )]
Assuming T? = 0°C:
16 = R? (1 + 15α)
20 = R? (1 + 100α)
Dividing the equations:
16/20 = (1+15α)/ (1+100α)
16 + 1600α = 20 + 300α
1300α = 4
α = 4/1300 ≈ 0.003 °C? ¹

As b > a
The magnetic field inside the wire (rR) is B = µ? I/ (2πr).
For wire with radius a, B increases linearly to r=a, then decreases. For wire with radius b, B increases linearly to r=b, then decreases. Since a⇒ B? > B?
B? = µ? I / 2πa
B? = µ? I / 2πb
(Note: The question is likely asking for the graph representation, which is option A based on the formulas.)

dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.

P (at least one head) = 1 - P (no heads) = 1 - (1/2)? ≥ 0.9.
0.1 ≥ (1/2)?
10 ≤ 2?
n=3, 2³=8. n=4, 2? =16.
Minimum value of n is 4.

(a+3b). (7a-5b) = 7|a|² - 5ab + 21ab - 15|b|² = 7|a|²+16ab-15|b|²=0.
(a-4b). (7a-2b) = 7|a|² - 2ab - 28ab + 8|b|² = 7|a|²-30ab+8|b|²=0.
Subtracting: 46ab - 23|b|² = 0 ⇒ 2ab = |b|².
Substituting: 7|a|² + 8|b|² - 15|b|² = 0 ⇒ 7|a|² = 7|b|² ⇒ |a|=|b|.
cosθ = ab/ (|a|b|) = ab/|b|² = (1/2)|b|²/|b|² = 1/2.
θ = 60°.

log? (18x-x²-77)>0 ⇒ 18x-x²-77>1 ⇒ x²-18x+78<0. Roots are 93.
log? (.)>0 ⇒ log? (.)>1 ⇒ 18x-x²-77>3 ⇒ x²-18x+80<0 (x-8) (x-10)<0.
8Domain is (8,10). a=8, b=10.
I = ∫? ¹? sin³x/ (sin³x+sin³ (18-x)dx. Using King's property.
I = ∫? ¹? sin³ (18-x)/ (sin³ (18-x)+sin³x)dx.
2I = ∫? ¹? dx = 2. I=1.

f (1)=1.
f (4)=f (2)²=1 or 4.
f (6)=f (2)f (3).
Possible functions determined by values at primes: f (2), f (3), f (5), f (7).
f (2) can be 1 or 2. f (3) can be 1 or 3. f (5)=1,5. f (7)=1,7.
If f (m)=m, f (mn)=mn. One function. f (x)=1 is another.
What if f (2)=1, f (3)=3? f (6)=3.

f (x) is discontinuous at integers x=1,2,3. P= {1,2,3}.
f (x) is not differentiable at integers and where x- [x]=1+ [x]-x ⇒ 2 (x- [x])=1 ⇒ {x}=1/2.
So at x=0.5, 1, 1.5, 2, 2.5.
Q= {0.5, 1, 1.5, 2, 2.5}. Sum of elements is not asked.
Number of elements in P=3, in Q=5. Sum = 8.
Let's check the solution. Q= {1/2, 1, 3/2, 5/2}.
The sum of number of elements: 3+5=8.